There are 5 essentially different ways to distribute the black balls. In each case I'll count the essentially different ways of distributing the white balls.
4, 0, 0, 0
The possible distinguishable ways of distributing the white balls are: 2-0-0-0, 1-1-0-0, 0-2-0-0 and 0-1-1-0. So 4 is the number.
3, 1, 0, 0
Here we can do it like this: 2-0-0-0, 1-1-0-0, 1-0-1-0, 0-2-0-0, 0-1-1-0, 0-0-2-0, 0-0-1-1. So 7.
2, 2, 0, 0
Once again: 2-0-0-0, 1-1-0-0, 1-0-1-0, 0-0-2-0, 0-0-1-1. 5 ways.
2, 1, 1, 0
And again: 2-0-0-0, 1-1-0-0, 1-0-0-1, 0-2-0-0, 0-1-1-0, 0-1-0-1, 0-0-0-2. 7 ways.
1, 1, 1, 1
Lastly: 2-0-0-0, 1-1-0-0. 2 ways.
In total, $4+7+5+7+2 = 25$ ways to distribute the balls.
We assume the balls are distinguishable, that is, have distinct labels.
Your calculation for the first problem, and the reasoning that led to it, are correct.
For the second problem, an analysis might go as follows. The reds can be placed in $2^2$ different ways. For each such way, the blues can be placed in $2^3$ different ways. And for each way of placing reds and blues, the greens can be placed in $2^2$ different ways.
The third problem, is, as you point out, straight Stirling, or, I would prefer to say, Inclusion/Exclusion.
The fourth is rather simple, there are only $3$ abstract balls. This one is a derangements problem, but one so small that applying derangements machinery is unreasonable. But if we had boxes of $10$ different colours, and balls of these colours, derangements would give us the answer.
Best Answer
For part (a), these are (unordered) partitions of $4$ into $3$ parts where parts can be $0.$ To ensure that you hit all possibilities, tackle three cases: exactly $2,1,$ or $0$ parts with nothing. There cannot be three parts with nothing. If exactly $2$ parts have nothing, the only possibility is $4+0+0.$ If exactly $1$ part has nothing, the possibilities are $3+1+0$ and $2+2+0.$ If exactly $0$ parts have nothing, then each part has at least $1.$ This means the remaining one gets assigned somewhere and the only possibility is $2+1+1.$ So the answer is $4.$
For the second part, treat white balls and black balls separately. Doing either one will let you square it to get the answer thanks to the multiplication principle. For white balls, there are $$\binom{2+3-1}{3-1}=\binom{4}{2}=6$$ possibilities, so the answer is $6^2=36.$ This is using sticks and stones.