Rather than just memorizing what formula to use and when, I find it helpful to think about what the formula is doing and why it works. If I wanted to know how many three digits numbers you can make, my line of thought would be something like this:
The first digit can't be 0, but it can be any other digit. So I have nine possibilities. For each of those nine possibilities, the second digit can be anything, including 0. That gives me 9*10 or 90 possibilities. And for each of those, the third digit can be anything. The final result is 9*10*10 or 900 possible three digit numbers.
Now, let's apply similar logic to your problem. You have to make a three-digit number using only 0, 2, 5, 6, and 9. So, noting that the first digit still can't be zero, how many possibilities are there for the first digit? And how many possibilities are there for the second and third digits? Now how many possibilities overall? Then for the second and third parts, figuring out how many of those possibilities are even/odd, consider what it means for a number to be odd. Given your limited selection of digits, what must the last digit be in order for a number to be odd? What does that suggest about how many odd numbers there are? And once you know that, you can quickly figure out how many even numbers there are too.
