Since we know the series lasted only five games, there are four possible outcomes:
$$
1. LWWWW\\
2. WLWWW\\
3. WWLWW\\
4. WWWLW
$$
Thus, the probability that they Team A lost game one is $1/4$.
This seems like a straightforward problem in conditional probability. Let $W_1$ be the event of winning the first game, $W_2$ be the event of winning the second game, and $F_b, F_i, F_m$ be the events of facing a beginner, an intermediate player, and a master.
We are given the unconditional probabilities $P(F_b) = P(F_i) = P(F_m) = \frac13$, and the conditional probabilities
$$
P(W_1 \mid F_b) = P(W_2 \mid F_b) = \frac{9}{10}
$$
$$
P(W_1 \mid F_i) = P(W_2 \mid F_i) = \frac{5}{10}
$$
$$
P(W_1 \mid F_m) = P(W_2 \mid F_m) = \frac{3}{10}
$$
We are asked to find $P(W_2 \mid W_1)$. You have already observed that $P(W_1) = \frac{17}{30}$. From Bayes's theorem, we can write
$$
P(F_b \mid W_1) = \frac{P(W_1 \mid F_b)P(F_b)}{P(W_1)} = \frac{9}{17}
$$
$$
P(F_i \mid W_1) = \frac{P(W_1 \mid F_i)P(F_i)}{P(W_1)} = \frac{5}{17}
$$
$$
P(F_m \mid W_1) = \frac{P(W_1 \mid F_m)P(F_m)}{P(W_1)} = \frac{3}{17}
$$
By total probability, we have
\begin{align}
P(W_2 \mid W_1) & = P(W_2 \mid W_1, F_b) P(W_1 \mid F_b) \\
& + P(W_2 \mid W_1, F_i) P(W_1 \mid F_i) \\
& + P(W_2 \mid W_1, F_m) P(W_1 \mid F_m)
\end{align}
Because $W_1$ and $W_2$ are conditionally independent, given the same opponent, we can rewrite this as
\begin{align}
P(W_2 \mid W_1) & = P(W_2 \mid F_b) P(W_1 \mid F_b) \\
& + P(W_2 \mid F_i) P(W_1 \mid F_i) \\
& + P(W_2 \mid F_m) P(W_1 \mid F_m) \\
& = \frac{9}{10} \times \frac{9}{17}
+ \frac{5}{10} \times \frac{5}{17}
+ \frac{3}{10} \times \frac{3}{17} \\
& = \frac{115}{170} = \frac{23}{34}
\end{align}
P.S. I realize it's just a framing device, but I can't believe a single chess player could win $90$ percent against a beginner, $50$ percent against an intermediate player, and $30$ percent against a master.
Best Answer
I dislike working with odds as a matter of principle, so all my calculations will be in terms of probability. We can rephrase things in terms of odds at the end.
We try then to calculate the probability that $A$ wins at least $2$ games out of $3$ as well as the probability that $A$ loses the first $2$ games against $B$.
As you correctly noted, if the odds in favor of $A$ winning against $B$ is $3:2$ that implies that the probability that $A$ wins a game is $\frac{3}{3+2}=\frac{3}{5}$.
To win at least two games out of three is equivalent to winning exactly two games or winning exactly three games out of three. The probabilities for these to occur can be found via the binomial distribution.
The probability then of winning at least two games is $\binom{3}{2}\left(\frac{3}{5}\right)^2\left(\frac{2}{5}\right)^1 + \binom{3}{3}\left(\frac{3}{5}\right)^3 = \dfrac{3\cdot 3^2\cdot 2 + 3^3}{5^3} = \frac{81}{125}$
The odds then are $81:(125-81)$ or rather $81:44$.
The second part of the problem is similarly calculated.