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 $n$
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.
Best Answer
The number of ways of placing $n$ labelled balls in $k$ unlabelled boxes with repetition allowed is $\left\{n\atop k\right\}$, a Stirling number of the second kind. There is a rather ugly explicit formula for them:
$$\left\{n\atop k\right\}=\frac1{k!}\sum_i(-1)^{k-i}\binom{k}ii^n\;.$$
They also satisfy a rather nice Pascal-like recurrence:
$$\left\{n\atop k\right\}=k\left\{n-1\atop k\right\}+\left\{n-1\atop{k-1}\right\}\;,$$
with initial condition $$\left\{n\atop 0\right\}=\left\{0\atop n\right\}=[n=0]\;,$$
where $[n=0]$ is an Iverson bracket. The explanation here of the recurrence is concise but reasonably clear.
If the balls are also unlabelled, you are in effect looking at partitions of $n$ into $k$ parts. The number of such partitions is sometimes denoted by $P(n,k)$ and satisfies the recurrence
$$P(n,k)=P(n-1,k-1)+P(n-k,k)$$
with initial conditions $P(n,k)=0$ for $k>n$, $P(n,n)=1$, and $P(n,0)=[n=0]$. The triangle of these numbers is OEIS A008284, and you’ll find more information and references there.
If you don’t allow repetition, clearly you must have $n\le k$, and since you can’t tell one box from another, there is only one way to distribute the balls, whether they’re labelled or not: $n$ boxes each containing a ball, and $n-k$ empty boxes.