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.
It is 3: the partitions of 5 into $\leq 2 $ parts. These are pretty much just the tuples you have listed: (5), (4,1), (3,2).
Best Answer
You are looking for the ways for partitioning a set with $6$ elements in three subsets. If you allow empty subsets, the answer is given by $${6\brace 1}+{6\brace 2}+{6\brace 3} = 122$$ otherwise it is given by ${6\brace 3}=90$, with ${n\brace k}$ being a Stirling number of the second kind.