We assume the balls are distinguishable, that is, have distinct labels.
Your calculation for the first problem, and the reasoning that led to it, are correct.
For the second problem, an analysis might go as follows. The reds can be placed in $2^2$ different ways. For each such way, the blues can be placed in $2^3$ different ways. And for each way of placing reds and blues, the greens can be placed in $2^2$ different ways.
The third problem, is, as you point out, straight Stirling, or, I would prefer to say, Inclusion/Exclusion.
The fourth is rather simple, there are only $3$ abstract balls. This one is a derangements problem, but one so small that applying derangements machinery is unreasonable. But if we had boxes of $10$ different colours, and balls of these colours, derangements would give us the answer.
$\begin{Bmatrix}
n\\
k
\end{Bmatrix}$ is the number of ways to distribute $n$ distinct objects into $k$ identical boxes with no empty boxes allowed/each box containing at least $1$ object. We will first consider distributions into $k$ distinct boxes.
The total number of ways to distribute $n$ distinct objects into $k$ distinct boxes with empty boxes allowed is: $$k^n$$
Let $A_i$ be the number of ways to perform the distribution with the condition that box $i$ must be an empty box. There are no restrictions on the other boxes. (This is equivalent to distributing the $n$ objects into the remaining $k-1$ boxes, empty boxes allowed.)
For example $A_1$ is the number of ways to distribute the objects into the $k$ boxes with box 1 being empty and is equivalent to distributing the objects into the remaining $k-1$ boxes, empty boxes allowed.
$$\implies |A_1| = (k-1)^n $$
Now we notice that $|A_1|=|A_2|=|A_3|=\dots$ since it doesn't matter which box is empty. (We are always distributing into the remaining $k-1$ boxes).
We know there are $\binom{k}{1}$ possible $A_i $
$$\sum_{i=1}^k|A_i| = \binom{k}{1}(k-1)^n$$
Using a similar argument, $|A_i\cap A_j|=(k-2)^n$ because it involves distribution into $k-2$ remaining boxes. Also, using the fact that $|A_1\cap A_2|=|A_1\cap A_3|=|A_1\cap A_4|...$ and that there are $\binom{k}{2}$ such terms,
$$\sum_{1\leq i,j\leq k}|A_i\cap A_j| = \binom{k}{2}(k-2)^n$$
On a similar note, we try extending this to when $l$ boxes must be empty:
$$\sum_{1\leq i,j\dots l\leq k}|A_i\cap A_j\cap \dots\cap A_l| = \binom{k}{l}(k-l)^n$$
$|A_1\cup A_2\cup \dots \cup A_k|$ is the number of ways to distribute $n$ distinct objects into $k$ distinct boxes with at least one empty box.
We are trying to find the distribution with no empty boxes: $$|\overline{A_1\cup A_2\cup \dots \cup A_k}|=k^n -|A_1\cup A_2\cup \dots \cup A_k|$$
By the principle of inclusion and exclusion,
$$=k^n-(|A_1|+|A_2|+|A_3|+...)+(|A_1\cap A_2|+|A_1\cap A_3|+|A_1\cap A_4|+...)-...$$
$$=k^n-\binom{k}{1}(k-1)^n+\binom{k}{2}(k-2)^n-\binom{k}{3}(k-3)^n+\dots +(-1)^l\binom{k}{j}(k-j)^n$$
$$=\sum_{l=0}^{k}(-1)^l\binom{k}{l}(k-l)^n$$
Let $j=k-l$, when $l=0$, $j=k$. When $l=k$, $j=0$.
$$=\sum_{j=k}^{0}(-1)^{k-j}\binom{k}{k-(k-j)}j^n (\text{this is technically wrong notation})$$
This is just the sum backwards. So we have:
$$=\sum_{j=0}^{k}(-1)^{k-j}\binom{k}{j}j^n$$
Now we divide by $k!$ since we are dealing with identical boxes:
$$\begin{Bmatrix}
n\\
k
\end{Bmatrix}
=\frac{1}{k!}\sum_{j=0}^{k}(-1)^{k-j}\binom{k}{j}j^n$$
Best Answer
There are actually lots of interesting uses of the column sums of the Stirling numbers of the second kind $\left\{ n \atop k\right\}$. The ordinary and exponential generating functions both have nice forms (nicer than those for the row sums): $$\sum_{n=k}^{\infty} \left\{ n \atop k\right\} x^n = x^k \prod_{j=1}^k \frac{1}{1-jx},$$ $$\sum_{n=k}^{\infty} \left\{ n \atop k\right\} \frac{x^n}{n!} = \frac{(e^x-1)^k}{k!}.$$ These can be used to prove various identities involving $\left\{ n \atop k\right\}$.
Multiplying $\left\{ n \atop k\right\}$ by certain other interesting numbers before summing also yields some nice identities, such as these: $$\sum_{r=k}^n \binom{n}{r} \left\{ r \atop k\right\} = \left\{ n+1 \atop k+1\right\},$$ $$\sum_{r=k}^n k^{n-r} \left\{ r-1 \atop k-1\right\} = \left\{ n \atop k\right\},$$ $$\sum_{r=k}^n \left[ n \atop r\right] \left\{ r \atop k\right\} = \binom{n}{k} \frac{(n-1)!}{(k-1)!} = L(n,k),$$ where $L(n,k)$ is a Lah number. (Lah numbers also show up in the answer to this recent question: $n$th derivative of $e^{1/x}$).
Column $k$ is also used to convert reciprocal falling factorial powers $x^{\underline{-k}}$ to ordinary reciprocal powers $x^{-n}$ via $$x^{\underline{-k}} = \sum_{n=k}^{\infty} (-1)^{n-k} \left\{ n \atop k\right\} \frac{1}{x^n},$$ in a sort of inverse to the fact that row $k$ is used to convert ordinary powers $x^n$ to falling powers $x^{\underline{k}}$.
A good reference is Chapter 8 of Charalambides' Enumerative Combinatorics.