Be careful, in the Chinese Remainder Theorem you must are relatively prime numbers, which 3 and 3 are not. Otherwise, it would mean than being divisible by 3 is the same as being divisible by 9...

Then I can't see a quicker solution than looking at the 9 cases (it is pretty quick though, since the remainder modulo 9 is just the sum of the digits). I think 1, 2 and 6 are solutions.

For the third one, using the CRS is correct however, since 5 and 9 have no common divisor except 1. What you get is that the solutions are the integers congruent to 1, 2 or 6 modulo 9, and to 1 or 3 modulo 5. This makes 3x2=6 possibilities of congruence systems. The CRS gives you that the solutions of these systems are the integers congruent to any particular solution modulo 45. For these particular solutions, I get 1 (congruent to 1 mod 5 and 1 mod 9), 11 (1 and 2), 6 (1 and 6), 28 (3 and 1), 38 (3 and 2) and 33 (3 and 6).