diff --git a/sql/task1.sql b/sql/task1.sql
index 90de336ca..f9ff77831 100644
--- a/sql/task1.sql
+++ b/sql/task1.sql
@@ -1,14 +1,71 @@
 -- Problem 1: Retrieve all products in the Sports category
 -- Write an SQL query to retrieve all products in a specific category.
 
+-- Select the name from products to be outputted. Category id is related in Categories and Products so join them
+-- Looking for the category id that is equal to 8
+SELECT 
+    P.product_name
+FROM 
+    Products P
+INNER JOIN 
+    Categories C ON P.category_id = C.category_id
+WHERE 
+    C.category_id = 8;
+
+
 -- Problem 2: Retrieve the total number of orders for each user
 -- Write an SQL query to retrieve the total number of orders for each user.
 -- The result should include the user ID, username, and the total number of orders.
 
+--Need User ID, Username and keep count of the number of orders for each user ID
+-- Joining user_id in both orders and users
+SELECT
+    U.user_id,
+    U.username,
+    COUNT(O.order_id) AS total_orders
+FROM
+    Users U
+LEFT JOIN
+    Orders O ON U.user_id = O.user_id
+GROUP BY
+    U.user_id, U.username;
+
+
 -- Problem 3: Retrieve the average rating for each product
 -- Write an SQL query to retrieve the average rating for each product.
 -- The result should include the product ID, product name, and the average rating.
 
+--Need to Average the ratings
+--Join the product_id from reviews to products
+SELECT
+    P.product_id,
+    P.product_name,
+    AVG(R.rating) AS Avg_Ratings
+FROM 
+    Products P
+JOIN
+    Reviews R ON P.product_id = R.product_id
+GROUP BY   
+    P.product_id, P.Product_name;
+
 -- Problem 4: Retrieve the top 5 users with the highest total amount spent on orders
 -- Write an SQL query to retrieve the top 5 users with the highest total amount spent on orders.
 -- The result should include the user ID, username, and the total amount spent.
+
+--LIMIT 5 at the end for the top 5 users
+--Need the SUM of the total amount in Orders
+--Join user_id from users and orders to have corresponding users to orders
+--Filter the amount spent in a descending order from most spent to least amount spent
+SELECT
+    U.user_id,
+    U.username,
+    SUM(O.total_amount) AS total_amount_spent
+FROM
+    Users U
+INNER JOIN
+    Orders O ON U.user_id = O.user_id
+GROUP BY
+    U.user_id, U.username
+ORDER BY
+    total_amount_spent DESC
+LIMIT 5;
\ No newline at end of file
diff --git a/sql/task2.sql b/sql/task2.sql
index ad2596731..e170de38a 100644
--- a/sql/task2.sql
+++ b/sql/task2.sql
@@ -3,17 +3,102 @@
 -- The result should include the product ID, product name, and the average rating.
 -- Hint: You may need to use subqueries or common table expressions (CTEs) to solve this problem.
 
+--Same idea as problem 3 but order the average rating in descending order from highest avg rating to lowest avg rating 
+SELECT
+    P.product_id,
+    P.product_name,
+    AVG(R.rating) AS average_rating
+FROM
+    Products P
+LEFT JOIN
+    Reviews R ON P.product_id = R.product_id
+GROUP BY
+    P.product_id, P.product_name
+ORDER BY
+    average_rating DESC;
+
+
 -- Problem 6: Retrieve the users who have made at least one order in each category
 -- Write an SQL query to retrieve the users who have made at least one order in each category.
 -- The result should include the user ID and username.
 -- Hint: You may need to use subqueries or joins to solve this problem.
 
+--Join Users, Orders, Order Items, Products and Categories tables together
+--Use subquery in WHERE to filter users who have made at least one order in each category by checking
+-- counts of distinct categories for each users
+--HAVING ensures the count for distinct categories matches the number of distinct categories in the Categories table
+SELECT DISTINCT 
+    U.user_id,
+    U.username,
+FROM 
+    Users U
+JOIN 
+    Orders O ON U.user_id = O.user_id
+JOIN 
+    Order_Items OI ON O.order_id = OI.order_id
+JOIN 
+    Products P ON OI.product_id = P.product_id
+JOIN 
+    Categories C ON P.category_id = C.category_id
+WHERE 
+    U.user_id IN (    
+        SELECT DISTINCT 
+            U2.user_id
+        FROM 
+            Users U2
+        JOIN 
+            Orders O2 ON U2.user_id = O2.user_id
+        JOIN 
+            Order_Items OI2 ON O2.order_id = OI2.order_id
+        JOIN 
+            Products P2 ON OI2.product_id = P2.product_id
+        JOIN 
+            Categories C2 ON P2.category_id = C2.category_id
+        GROUP BY 
+            U2.user_id, C2.category_id
+        HAVING 
+            COUNT(DISTINCT C2.category_id) = (SELECT COUNT(DISTINCT category_id)FROM Categories)
+);
+
 -- Problem 7: Retrieve the products that have not received any reviews
 -- Write an SQL query to retrieve the products that have not received any reviews.
 -- The result should include the product ID and product name.
 -- Hint: You may need to use subqueries or left joins to solve this problem.
 
+--Join product_id from reviews and products
+--only output the product_id and name if the review_id is NULL (empty/not exist) for any product_id in Products
+SELECT 
+    P.product_id, P.product_name
+FROM 
+    Products P
+LEFT JOIN 
+    Reviews R ON P.product_id = R.product_id
+WHERE 
+    R.review_id IS NULL;
+
 -- Problem 8: Retrieve the users who have made consecutive orders on consecutive days
 -- Write an SQL query to retrieve the users who have made consecutive orders on consecutive days.
 -- The result should include the user ID and username.
--- Hint: You may need to use subqueries or window functions to solve this problem.
\ No newline at end of file
+-- Hint: You may need to use subqueries or window functions to solve this problem.
+
+--Using CTE to create a new table for OrderedOrders
+--Use LEAD() window function to get next order_date for each row from the result set
+--Partition result set by user id so the lead function is applied independently to each user and order the rows by order date within each partition
+--Assign the above results into a new column called next_order_date
+--WHERE filters wanted results that have a difference between consecutive orders of 1 day or next order date is NULL
+WITH OrderedOrders AS (
+    SELECT
+        user_id,
+        order_id,
+        order_date,
+        LEAD(order_date) OVER (PARTITION BY user_id ORDER BY order_date) AS next_order_date
+    FROM
+        Orders
+)
+SELECT DISTINCT U.user_id, U.username
+FROM
+    Users U
+JOIN
+    OrderedOrders O ON U.user_id = O.user_id
+WHERE
+    DATEDIFF(next_order_date, order_date) = 1 OR next_order_date IS NULL;
diff --git a/sql/task3.sql b/sql/task3.sql
index f078a9439..252945bcf 100644
--- a/sql/task3.sql
+++ b/sql/task3.sql
@@ -3,17 +3,96 @@
 -- The result should include the category ID, category name, and the total sales amount.
 -- Hint: You may need to use subqueries, joins, and aggregate functions to solve this problem.
 
+--Use CTE called Category_Sales to include category_id, category_name and the sum of total_amount which would be the total_sales
+--join all corresponding ids (such as category, product and order) from categories with products, order_items and orders
+WITH Category_Sales AS (
+    SELECT
+        C.category_id,
+        C.category_name,
+        SUM(O.total_amount) AS total_sales
+    FROM
+        Categories C
+    JOIN
+        Products P ON C.category_id = P.category_id
+    JOIN
+        Order_Items OI ON P.product_id = OI.product_id
+    JOIN
+        Orders O ON OI.order_id = O.order_id
+    GROUP BY
+        C.category_id, C.category_name
+)
+
+--Select the category id, name and total sales as output from the new table and order it by having the highest total sales at the top
+SELECT
+    category_id,
+    category_name,
+    total_sales
+FROM
+    Category_Sales
+ORDER BY
+    total_sales DESC
+LIMIT 3;
+
 -- Problem 10: Retrieve the users who have placed orders for all products in the Toys & Games
 -- Write an SQL query to retrieve the users who have placed orders for all products in the Toys & Games
 -- The result should include the user ID and username.
 -- Hint: You may need to use subqueries, joins, and aggregate functions to solve this problem.
 
+--Filter products to those in "Toys & Games"
+--HAVING used to make sure the count of distinct product IDs for each user is = total number of products in "Toys & Games" category
+SELECT
+    U.user_id,
+    U.username
+FROM
+    Users U
+JOIN
+    Orders O ON U.user_id = O.user_id
+JOIN
+    Order_Items OI ON O.order_id = OI.order_id
+JOIN
+    Products P ON OI.product_id = P.product_id
+WHERE
+    P.category_id = 5
+GROUP BY
+    U.user_id, U.username
+HAVING
+    COUNT(DISTINCT P.product_id) = (
+        SELECT COUNT(*)
+        FROM Products
+        WHERE category_id = 5
+    );
+
 -- Problem 11: Retrieve the products that have the highest price within each category
 -- Write an SQL query to retrieve the products that have the highest price within each category.
 -- The result should include the product ID, product name, category ID, and price.
 -- Hint: You may need to use subqueries, joins, and window functions to solve this problem.
 
+--Make a CTE called Ranked_Products and use ROW_NUMBER() function to assign ranks to each product within its category based on prices in desc order
+--ORDER BY in OVER clause ensures that the ranking is reset for each category
+--Main query selects products from the CTW where row_num will equal 1 -> meaning it'll select the products has the highest price in each category 
+WITH Ranked_Products AS (
+    SELECT
+        product_id,
+        product_name,
+        category_id,
+        price,
+        ROW_NUMBER() OVER (PARTITION BY category_id ORDER BY price DESC) AS row_num
+    FROM
+        Products
+)
+
+SELECT
+    product_id,
+    product_name,
+    category_id,
+    price 
+FROM
+    Ranked_Products 
+WHERE
+    row_num = 1;
+
 -- Problem 12: Retrieve the users who have placed orders on consecutive days for at least 3 days
 -- Write an SQL query to retrieve the users who have placed orders on consecutive days for at least 3 days.
 -- The result should include the user ID and username.
 -- Hint: You may need to use subqueries, joins, and window functions to solve this problem.
+