Let's try to break up each parts into subpart:
Balls and boxes are labeled, so $1111100\neq 1100111$ where $1$ is a filled box and $0$ an empty one, and $12345\neq12435$ where I labeled each ball.
1a. Choose which box you want to fill. They all contain exactly one ball.
1b. The problem is now a standard permutation problem
Boxes aren't labeled so there really is just one way to do this, putting $5$ balls into $5$ bins.
Balls are not labeled here, so this is part a) minus ii). Just choose which bin to fill
Same as b), not much to be done here.
Now to answer you question more precisely, let us look at a broader example. Imagine I have $4$ balls and $3$ bins, but no restriction on the number of balls each bin can hold.
If both are labeled, then just pick a bin for each balls. This give $3^4$ ways since each ball can go in any of the $3$ box.
If balls are unlabeled both boxes are différent, then this is classic Stars and bars counting technique.
The harder part comes when boxes are unlabeled. First if the balls are labeled then there are $4$ ways the ball can be distributed, namely
$$
\{\{4,0,0\},\{3,1,0\},\{2,2,0\},\{1,1,2\}\}
$$
Now $\{4,0,0\}$ has only $1$ possible configuration and $\{3,1,0\}$ has $4$, i.e. choose the ball that is alone in its bin. The remaining two situations each have $6$ possible configurations, that is choose $2$ balls to be together from the $4$, the others being set depending on a $1-1$ configuration or $2-0$. This gives a total of $15$ configurations.
In general, this is given by Bell number. If boxes can't be empty, you'll want to look up Stirling number of the second kind.
If everything is unlabeled, then this is given (in general, $n$ balls into $k$ non-empty bins) by the number of $k$-element partitions of $n$
OK, so I found a useful resource that helped me understand (I think) how to apply the idea of partitioning to this problem.
Before starting, I should say that we will assume each box can contain all 8 objects, no objects, or any integer in between. Other answers may use other assumptions.
Basically:
Definition: P(k, i) is the number of partitions of k into i parts.
In this case P(8,i), where i $\leq 8$.
P(k) is the number of partitions of k into any number of parts.
In this case we want P(8).
So we need to enumerate all the possible combinations:
P(8,1) = 1 (i.e. 8 objects in one box, all others empty.)
P(8,2) = 4 [(6,2), (7,1), (5,3), (4,4)]
P(8,3) = 5 [(5,2,1), (3,3,2), (4,2,2), (4,3,1), (6,1,1)]
P(8,4) = 5 [(2,2,2,2), (4,2,1,1), (3,2,2,1), (5,1,1,1), (3,3,1,1)]
P(8,5) = 3 [(4,1,1,1,1), (3,2,1,1,1), (2,2,2,1,1)]
P(8,6) = 2 [(3,1,1,1,1,1), (2,2,1,1,1,1)]
P(8,7) = 1 [(2,1,1,1,1,1,1)]
P(8,8) = 1 [(1,1,1,1,1,1,1,1)]
P(8) = $\sum{P(8,i)}$ = 22
Therefore the answer is 22.
EDIT: "A recurrence relation is part(n,k) = part(n-1,k-1) + part(n-k,k)"
This is from: http://2000clicks.com/mathhelp/CountingObjectsInBoxes.aspx
The site contains a link to the proof.
Best Answer
Here is how I put it: Choose $n_1$ boxes out of $k$ boxes and place all the $n_1$ objects into this selection. Among the remaining $k-n_1$ boxes, choose $n_2$ boxes and place all the $n_2$ boxes in this selection. So, the answer would be ${k \choose n_1}{k-n_1\choose n_2}$, which matches to your answer.
Probably similar to what is given by User58220