What a fun question! For (i) there are 2^n players. We want 2 of them, A and B, to play in the first round. Let's focus on one of our selected players, player A. A is absolutely 100% going to play SOMEONE in the first round, each player being an equally likely opponent. If there were 2^n players originally...and we take A out of that pool, there are now (2^n) - 1 players left for A to play against. B is one of those players. So the probability that A will play against B in the first round is 1/((2^n) - 1).
For (ii), there are 3 independent events that must occur: 1) they must be on opposite sides of the bracket, 2) A must win n-1 games and 3) B must win n-1 games.
So.....first they would have to be on opposite sides of the bracket initially. This is gonna get messy! There are 2^n players. Half are going to be on each side of the bracket. So (2^n)/2 = 2^(n-1) players will be on each side. (2^n means you have n "2s"....so if you divide by a 2, you now have 1 less than n "2s".) Let's start again by focusing on A. A is on one side of the bracket. This leaves (2^n) - 1 "slots" to fill out for the entire bracket, (see (i)). If B is going to be on the "other" side, there are 2^(n-1) slots over there out of (2^n) - 1 slots all around. So the probability that B is on the other side is [2^(n-1)]/[(2^n) - 1]. That's the first independent "event" that needs to happen!
Each player must now win all rounds leading up to the final. The problem stated that there were n rounds, so each player needs to win n-1 single-elimination games. Since each player is equally likely to win, that means each has a 1/2 chance to win each game. So player A, for example, has a (1/2)^(n-1) or 1/(2^(n-1)) of reaching the final. So does B.
Each of these events...the bracket arrangement, A winning through to the final, and B doing likewise are independent events. We need all 3 to happen....so we will multiply their probabilities. That leaves us with:
[2^(n-1)] / [(2^n - 1)(2^(n-1))(2^(n-1))]. We see 2^(n-1) in both the numerator and denominator so we can divide to 1 leaving us with a final probability of: 1/ [(2^n - 1)(2^(n-1)].
(Yes we could distribute, but it wouldn't really simplify anything.....we'd still have 2 terms in the denominator, so why bother!?
I'm going to leave (iii) for another day!
