Grouping people into interchangeable teams At the end of Q14a below, what can it mean that "the two teams of size three become interchangeable"? If they become a single team of size 6, shouldn't the number of ways of choosing teams become $\binom{8}{2}\binom{6}{6}=\binom{8}{2}=28$?
Similarly, at the end of Q14b, why should it be that "if we remove the hats, we can reassign the team colours in $4!$ ways"? I can see how that would be the case if there were 4 people to assign to 4 teams each of size 1, but not with 8 people.
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To put it another way: [Q14(a)i] asks how many 8-letter words can be spelled with 2 R's, 3 G's and 3 B's; my conjecture as to the meaning of [Q14(a)ii] was to ask how many 8-letter words can be spelled with 2 R's and 6 G/B's; the interpretation I can't make sense of is to ask how many 8-letter words can be spelled with 2 R's, 3 G/B's and 3 more G/B's.
If we can't distinguish between them, in what sense are there two teams?
 A: Your illustrations [Q14(a)i] $R,R,\,B,B,B,\,G,G,G\,$ and [Q14(a)ii] $R,R,\,G,G,G,G,G,G$ don't correspond to the original problem.
A valid distillation (which doesn't under-count) is $$R_1,R_2,g_1,g_2,g_3,G_1,G_2,G_3$$ where an outsider attempting to discern team membership is $$\textbf{[Q14(a)i]}\text{ case-sensitive  vs. }\textbf{[Q14(a)ii]}\text{ case-insensitive.}$$
Crucially, the fact that external parties can't distinguish between teams $g$ and $G$ (due to case-insensitivity) does not negate the extant partition of the two teams; i.e., $g$ and $G$ are still separate teams. (If, on the contrary, their members intermix, [Q14(a)ii] would be distilled as $R_1,R_2,\,G_1,G_2,G_3,G_4,G_5,G_6\,$ as you'd originally (mis)interpreted.)
So for [Q14(a)ii]:

*

*There are $\binom{8}{2}$ ways to select the red team from the eight people, $\binom{6}{3}$ ways to independently select the green team from the remaining six people, and $\binom{3}{3}$ ways to independently select the blue team from the last three people.
However, to a colour-blind observer, the green and blue teams are indistinguishable, so we need to compensate for the over-counting by dividing by the number of ways $2!$ to arrange two different groups.
Therefore the total number of ways is $$\binom{8}{2}\binom{6}{3}\binom{3}{3}\div\left(2!\right)=280.$$


*Alternative framing: The number of ways to arrange $n$ objects where $p$ are identical of a kind, $q$ are identical of another kind,
etc. is $\displaystyle\frac{n!}{p!q!r!...},$ so the number of ways to group 8 people into teams of 2,3,3, where same-sized teams are indistinguishable, is $$\frac{8!}{2!3!3!}\div\left(2!\right)=280.$$
For [Q14(b)ii], by the same reasoning, the solution is $$\binom{8}{2}\binom{6}{2}\binom{4}{2}\binom{2}{2}\div\left(4!\right)=105=\frac{8!}{2!2!2!2!}\div\left(4!\right).$$
A: What they evidently intended by "removing the hats" is that they are still two teams with three people but we can no longer distinguish them as the blue team or the green team.  So we have the divide the number of arrangements by 2! to account for that.  Same with the second problem except it is dividing by 4!.
To see the same effect with smaller numbers, see how many ways there are to divide four unique coins into a pile of 2 and two piles of 1. There are 12 ways to do that if you put the coins into unique bowls, but only 6 if you put the coins into piles without the bowls. 
A: It is a case of distribution of distinct objects into identical groups.$$ $$
Let (mn) number of distinct objects be distributed to m numbers of identical groups such that each gets n objects is $$\frac{(mn)!}{(n!)^m.m!}$$
Groups can be identical if they are of same size. $$ $$
Hence in your question number of ways will be $$=\frac{8!}{2!.(3!)^2.2!}$$
