Although the problem doesn't say so, I think we have to assume that each game is won by one of the players, i.e. a draw is not a possibility (unlike in the Real World, where draws are quite common).
So let's assume each player has a 1/2 probability of winning each game, and let's also assume the outcomes are independent (the outcome of one game does not influence the outcome of the next). With these assumptions, the number of games a player wins in n games has a Binomial distribution with p=1/2 and n=n (what else?). Then the probability of x wins in n games is
So the probability of winning exactly 4 games out of 6 is
and the probability of winning exactly 5 games out of 9 is
I'll leave you to do the arithmetic, and maybe with this start you can work the other problems too.