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 $n$
Each box here can contain either 0 or 1 ball, and there are $y$ boxes. This fact can be represented using binomial coefficients. We can rephrase the question as:
$(1+t)(1+t)....(1+t)$ where $(1+t)$ has been multiplied $y$ times. What is the coefficient of $t^x$? Clearly the answer is $\binom {y}{x}$.
Here, $(1+t)$ has been used for each box, because the power of $t$ represents the number of balls in that box. Thus, multiplying them together gives all possible combinations.
Best Answer
Your answer is incorrect because when you do $m^n$, you are treating each particle separately by giving each one $m$ choices, essentially making them "distinguishable." To see this, let's say you give the first choice to particle $p_1$, second choice to $p_2$, and so on, all the way to $p_n$. Then you will see that something like $$p_1,p_2,p_3,p_4,\ldots,p_n,E,E,\dots,E$$ where $E$ represents empty boxes would be counted differently from $$p_2,p_1,p_3,p_4,\ldots,p_n,E,E,\ldots,E$$ when, in fact, they should be counted the same. To account for this, you would need actually need to do a method known as Stars and Bars, which you can look into further in the Wikipedia page linked.