To find the number of 4-digit numbers that can be formed using the digits 0, 2, 3, 5, and 8 without repetition, such that each number is not divisible by 4, we can use the concept of counting principles.

We can break down the problem into several cases:

Case 1: Numbers ending with 0
- Since the number has to be 4 digits, the first digit cannot be 0.
- Therefore, there are no numbers ending with 0 that satisfy the given condition.

Case 2: Numbers ending with 2
- The last digit is fixed as 2.
- The first digit cannot be 0 or 2, so there are 3 choices (3, 5, and 8) for the first digit.
- The remaining two digits can be any of the remaining 3 digits.
- Therefore, there are 3 choices for the first digit and 3 choices for the second digit, resulting in a total of 3x3 = 9 numbers.

Case 3: Numbers ending with 3
- Similar to Case 2, the last digit is fixed as 3.
- The first digit cannot be 0 or 3, so there are 3 choices (2, 5, and 8) for the first digit.
- The remaining two digits can be any of the remaining 3 digits.
- Therefore, there are 3 choices for the first digit and 3 choices for the second digit, resulting in a total of 3x3 = 9 numbers.

Case 4: Numbers ending with 5
- The last digit is fixed as 5.
- The first digit cannot be 0, 2, or 5, so there are 2 choices (3 and 8) for the first digit.
- The remaining two digits can be any of the remaining 3 digits.
- Therefore, there are 2 choices for the first digit and 3 choices for the second digit, resulting in a total of 2x3 = 6 numbers.

Case 5: Numbers ending with 8
- The last digit is fixed as 8.
- The first digit cannot be 0 or 8, so there are 3 choices (2, 3, and 5) for the first digit.
- The remaining two digits can be any of the remaining 3 digits.
- Therefore, there are 3 choices for the first digit and 3 choices for the second digit, resulting in a total of 3x3 = 9 numbers.

Adding up the numbers from all the cases, we get a total of 0 + 9 + 9 + 6 + 9 = 33 numbers.

Therefore, the correct answer is option D, 66.