You are correct that the only way to win it in 6 games while losing two games at home is to lose the first two games, then win all four games that follow, so there is a single way of doing it.
To win in 7 games while losing two games at home, you can lose any two of the first, second, and sixth game (giving you $C(3,2)$), and lose one of the games on the road (giving your $C(3,1)$), and win the rest; so again, you are correct there.
However, $C(3,2) = C(3,1) = 3$, so $C(3,2)\times C(3,1) = 9$; the total is $10$, not $11$.
This problem is not done correctly. The fact that each team can win each game with equal probability does not mean that all final outcomes are equally likely. For example, there are only two ways in which the series can last 4 games, and each of them can occur with probability $(1/2)^4 = 1/16$. So there is only $1/8$ probability of the series lasting 4 games. For the series to last 5 games, we must have one of 21111, 12111, 11211, or 11121; or 12222, 21222, 22122, 22212; each of them has probability $(1/2)^5 = 1/32,$ so the probability is $1/4$, double the probability that the series will last 4 games.
For the series to last 6 games with team 1 winning, team 2 must win two of the first five games; there are $C(5,2) = 10$ ways of doing this; each one has probability $(1/2)^6 = 1/64$; same count for team 2 winning, so that gives $20/64 = 5/16$ probability of the series lasting 6 games.
For the series to last 7 games with team 1 winning, team 2 has to win three of the first six; there are $C(6,3) = 20$ ways of this happening, each with probability $(1/2)^7 = 1/128$; same for team 2 winning, so we get $40/128 = 5/16$ probability.
(Gut check: $(1/8) + (1/4) + (5/16) + (5/16) = 1$; and, with evenly matched teams, the most likely outcome is either a 6 game series or a 7 game series, which sounds reasonable).
Best Answer
The sequences that lead to a 4th win by A in game 6 are
WWWLLW 1
WWLWLW 2
WWLLWW 3
WLWLWW 4
WLWWLW 5
WLLWWW 6
LWWWLW 7
LWWLWW 8
LWLWWW 0
LLWWWW 10
These are 10 sequences because the 6th slot is fixed as a win and the arrangements of the other 5 slots is $^5$C$^3$ =5! /(3! 2!)=5 (4)/2 = 10.
The game H = home for team A and R = road for team A sequence for 6 games HHRRHRH Then assuming the Markov property these 10 probabilities are
p(1)= (.65)(.60)(.40)(.60)(.30)(.45)=0.012636 based on the stated conditional probabilties for wins and loss on home and road games after a win and after a loss and the probability of a win in the first game.
p(2)= (.65)(.60)(.60)(.45)(.40)(.45)=0.018954
and you can do the rest.