Might as well comment on this. Hopefully I don't make an error and add to confusion.
1) It is valid to consider 495 equally likely arrangements as has been done by others, noting that some of these arrangements are equivalent up to rotation. This means: If we consider all distinct arrangements up to rotation, we may have difficulty using these counts for probability because *they are not all equally likely* given a random placement of marbles.
Now, let's simplify a bit and consider them placed linearly. Using visual aids,
| B | B | B | B | B | B | B | B |
Considering just the linear case, there are C(9,4) = 126 arrangements in which no two red marbles are adjacent. But of course we must take out those that have marbles in both the first and last positions. There are C(7,2) = 21 of these. So in the end we get (126-21)/495 = 7/33, in agreement with some previous answers.
2) Here there are 5^6 distinct possibilities, all equally likely. I think we can count directly, rather than counting the opposite of what we want and taking the complement. We will use principle of inclusion-exclusion. Consider selecting any pair and making them choose each other. There are C(6,2) = 15 ways to choose the pair, and 1 way they can choose each other. For each of these 15 ways, we have 5^4 = 625 ways the other people can choose. But of course if we do 15*5^4 we count duplicates. So, select two pairs and make them choose each other. There are C(6,4) * C(4,2) / 2 = 45 ways to do this. And the other people can choose in 5^2 = 25 ways. Now finally let's make it so all three have chosen in pairs. There are 5*3*1 = 15 ways to break into pairs. So we have numerator
15*5^4 - (C(6,4) * C(4,2) * 5^2) + 5*3*1 = 8265
So we get probability 8265/5^6 = 1653/3125
which matches what bright s hell got.