Having trouble with my Computer Science task

[DeaD_EyE]

int("Guess the number that was generated from 1-10")

This will raise a ValueError because "Guess the number that was generated from 1-10" is not a valid int.

First get the user-input, which is a str, then convert it to an int.

user_input = input("Your number: ")
user_value = int(user_input)

# or in one line
user_value = int(input("Your number: "))

In line 3 you assign the random int to the name i.
Don't use characters for names. Use names, which are telling the programmer which object this could be.
For example, random_value is a good name.

In line 14 you use a bare elif without a condition. elif requires a condition. else does not require a condition. In line 24 you made it vice versa. else does not accept a condition. You wanted to use a bare else (because there are no more possibilities) or you could replace else with elif condition.

User input is always error-prone. A user could enter an invalid number, which int() can't convert to an int. This will raise a ValueError, which ends the program because you're not catching the exception.

And the last, but most important is the indentation.
Never use an editor, which indents tabulator instead of 4 spaces.
Never use tab stops in Python source code.
Use a better editor. VS Code, Atom, Code with Mu or other...
PyCharm is too heavy for beginners.

Here the code applied with all suggestions:

import random


def ask_user_for_number():
    """
    Asking the user to input a valid integer and return this value as int.
    Otherwise repeat the question.
    """
    while True:
        user_input = input("Guess the number that was generated from 1-10: ")

        # now the conversion to an int can fail
        # using exception handling to detect this
        try:
            user_value = int(user_input)
            # int("abc") -> ValueError
        except ValueError:
            print("Your input is not an integer:", user_input)
            # give the user a message and print also his wrong input
        else:
            # this block is only executed, if no excption happens
            # return user_value leaves the function context and the caller gets
            # user_value
            return user_value


count = 0
# use names which humans can understand
random_value = random.randint(1, 11)


while True:
    # here is the call
    guess = ask_user_for_number()

    if guess == random_value:
        # format string. Makes it easier to print text with variables.
        print(f"Success. The number is {random_value}")
        # but this won't stop the while True loop
        # just use break to stop the break out of the loop
        break
    elif random_value < guess:
        print("The number is smaller")
    else:
        # this is the last possible case
        print("The number is greater")

    # always increment
    # reduces code duplication
    # instead of count = count + 1
    # you can write count += 1
    count += 1


if count == 1:
    print("You got 4 stars")
elif count >= 2:
    print("You got 3 stars")
elif count == 0:
    print("you got 5 stars")

Hint: Use Python 3.10.0 because this version does have better description in exceptions, and sometimes you get hints that you've forgotten a colon.

Hint2: You can put the part on module level, where you print how many stars you got, in a function, which take one argument, which is the count of guesses.

Same code, but in a function:

def star_message(count):
    if count == 1:
        print("You got 4 stars")
    elif count >= 2:
        print("You got 3 stars")
    elif count == 0:
        print("you got 5 stars")

Then you can call this:

star_message(0)
star_message(1)
star_message(2)