diff --git a/solutions.sql b/solutions.sql
index d0eddcc..e974910 100644
--- a/solutions.sql
+++ b/solutions.sql
@@ -1 +1,118 @@
+-- 1. How many copies of the film Hunchback Impossible exist in the inventory system?
+SELECT 
+    COUNT(*) AS copies_count
+FROM 
+    inventory i
+JOIN 
+    film f ON i.film_id = f.film_id
+WHERE 
+    f.title = 'Hunchback Impossible';
+
+-- 2. List all films whose length is longer than the average of all the films.
+SELECT 
+    title, 
+    length
+FROM 
+    film
+WHERE 
+    length > (SELECT AVG(length) FROM film);
+
+-- 3. Use subqueries to display all actors who appear in the film Alone Trip.
+SELECT 
+    a.first_name,
+    a.last_name
+FROM 
+    actor a
+WHERE 
+    a.actor_id IN (SELECT fa.actor_id
+                   FROM film_actor fa
+                   JOIN film f ON fa.film_id = f.film_id
+                   WHERE f.title = 'Alone Trip');
+
+-- 4. Identify all movies categorized as family films.
+SELECT 
+    f.title
+FROM 
+    film f
+JOIN 
+    film_category fc ON f.film_id = fc.film_id
+JOIN 
+    category c ON fc.category_id = c.category_id
+WHERE 
+    c.name = 'Family';
+
+-- 5. Get name and email from customers from Canada using subqueries.
+SELECT 
+    first_name, 
+    last_name, 
+    email
+FROM 
+    customer
+WHERE 
+    address_id IN (SELECT address_id 
+                   FROM address
+                   WHERE city_id IN (SELECT city_id 
+                                     FROM city
+                                     WHERE country_id = (SELECT country_id 
+                                                         FROM country 
+                                                         WHERE country = 'Canada')));
+
+-- 5 (continued): Get name and email from customers from Canada using joins.
+SELECT 
+    c.first_name, 
+    c.last_name, 
+    c.email
+FROM 
+    customer c
+JOIN 
+    address a ON c.address_id = a.address_id
+JOIN 
+    city ci ON a.city_id = ci.city_id
+JOIN 
+    country co ON ci.country_id = co.country_id
+WHERE 
+    co.country = 'Canada';
+
+-- 6. Which are films starred by the most prolific actor?
+SELECT 
+    f.title
+FROM 
+    film f
+JOIN 
+    film_actor fa ON f.film_id = fa.film_id
+WHERE 
+    fa.actor_id = (SELECT actor_id
+                   FROM film_actor
+                   GROUP BY actor_id
+                   ORDER BY COUNT(film_id) DESC
+                   LIMIT 1);
+
+-- 7. Films rented by the most profitable customer.
+SELECT 
+    f.title
+FROM 
+    payment p
+JOIN 
+    rental r ON p.rental_id = r.rental_id
+JOIN 
+    inventory i ON r.inventory_id = i.inventory_id
+JOIN 
+    film f ON i.film_id = f.film_id
+WHERE 
+    p.customer_id = (SELECT customer_id 
+                     FROM payment 
+                     GROUP BY customer_id 
+                     ORDER BY SUM(amount) DESC 
+                     LIMIT 1);
+
+-- 8. Get the client_id and the total_amount_spent of those clients who spent more than the average of the total amount spent by each client.
+SELECT 
+    customer_id, 
+    SUM(amount) AS total_amount_spent
+FROM 
+    payment
+GROUP BY 
+    customer_id
+HAVING 
+    total_amount_spent > (SELECT AVG(total) FROM (SELECT SUM(amount) AS total FROM payment GROUP BY customer_id) AS avg_totals);
 -- Add you solution queries below: