I will assume that you know how to count the number of ways to distribute $n$ identical candies among $r$ children. Compute this number for the blue balls, call it $B$. Compute this number for the red balls. Call it $R$. For every way of distributing the blue balls, there are $R$ ways to distribute the red balls, so our total is $BR$.
If you do not know how to solve the general $n$ candy $r$ child problem, I suggest you look at the Wikipedia article on Stars and Bars. The answer turns out to be $\binom{n+r-1}{r-1}$. The reasoning is well-described in the Wikipedia article, and in several MSE answers.
For our particular very small numbers, $B$ and $R$ can also be computed by making careful lists.
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
If the balls and boxes are labelled, there are $\binom53$ ways to choose the three non-empty boxes. Suppose that we’ve chosen the three boxes that are to be non-empty. There are $3^5$ ways to put the $5$ balls into those boxes, but some of those ways leave one or more of the three boxes empty. To correct for this we use an inclusion-exclusion argument.
Call the three chosen boxes Box 1, Box 2, and Box 3. There are $2^5$ ways to put the $5$ balls into Boxes 2 and 3, so that Box 1 remains empty. Similarly, there are $2^5$ ways to put the balls into Boxes 1 and 3, and $2^5$ ways to put the balls into Boxes 2 and 3. Thus, out of the $3^5$ arrangements of the balls in the three boxes, $3\cdot2^5$ leave a box empty and should be subtracted from the total to leave $3^5-3\cdot2^5$. But the arrangement that puts all $5$ balls into Box 3 got removed twice, since it leaves both Box 1 and Box 2 empty, and the same goes for the other two arrangements that put all $5$ balls into one of the three boxes. These should have been deducted only once from the original $3^5$, so we must add the $3$ back in, and the final count is
$$3^5-3\cdot2^5+3=243-96+3=150\;.$$
If the balls and boxes are both unlabelled, we simply want the number of partitions of $5$ into $3$ parts. This problem is small enough to be done by brute force: there are just two, $5=3+1+1$ and $5=2+2+1$.