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.
I'll solve the case in which the balls are identical. This is not the nicest solution, but it works and I hope it is clear.
Case 1: First two boxes have exactly four balls.
If the first two boxes together contain four balls, then we can use stars and bars twice. The number of ways to place four balls in two boxes is the number of ways to arrange $4$ stars and $1$ bar, or $5$ choose $1$. The number of ways to place the remaining four balls in the other four boxes is the number of was to arrange $4$ stars and $3$ bars, or $7$ choose $3$. This gives
$$\binom{5}{1}\binom{7}{3}$$
Case 2: First two boxes have exactly three balls. With the same reasoning, we get
$$\binom{4}{1}\binom{8}{3}$$
Case 3: First two boxes have exactly two balls.
$$\binom{3}{1} \binom{9}{3}$$
Case 4: First two boxes have exactly one ball.
$$\binom{2}{1}\binom{10}{3}$$
Case 5: First two boxes have no balls.
$$\binom{1}{1}\binom{11}{3}$$
(This makes sense because this is just stars and bars applied to putting $8$ balls into $4$ boxes)
Our total is therefore
$$5\binom{7}{3}+4\binom{8}{3}+3\binom{9}{3}+2\binom{10}{3}+\binom{11}{3}=\boxed{1056}$$
Best Answer
1) Your answer is correct; for each ball, you can choose any box, and every choice is distinguishable at any time.
2) You want to distribute your 5 distinguishable balls into 3 indistinguishable boxes. Let $B(5,3)$ denote the number of ways in which this can be done into exactly 3 indistinguishable non-empty boxes, and use the recurrence relation $B(n,k)=B(n-1,k-1)+kB(n-1,k)$ with $B(n,1)=1$ and $B(n,n)=1$. You seek $B(5,1)+B(5,2)+B(5,3)$. For further reference, see Stirling number.
3) You want to separate your 5 indistinguishable balls into 3 distinguishable boxes. Note that every admissible separation uniquely corresponds to a permutation of the string $XXbbbbb$.
4) You want to partition your 5 indistinguishable balls into 3 indistinguishable boxes. Let $p(5,3)$ denote the number of ways in which this can be done into exactly 3 indistinguishable non-empty boxes, and use the recurrence relation $p(n,k)=p(n-1,k-1)+p(n-k,k)$ with $p(n,1)=1$ and $p(n,n)=1$. You seek $p(5,1)+p(5,2)+p(5,3)$. For further reference, see partition.