What are the odds of A winning 2 games given that the odds of A winning a game is 3:2 The odds in favor of A winning a game of chess against B are 3:2. If 3 games are to be played, what are the odds: 
(a) in favor of A winning at least 2 games out of 3,
(b) against A losing the first 2 games to B
It is known that the answer are:
(a) 81:44
(b) 21:4
I know this sentence The odds in favor of A winning a game of chess against B are 3:2. means:  out of 5 games, A will ideally win 3 times and lose 2 times. In other words,the probability of A to win a game is 3/(3+2) = 3/5 but I do not understand how the answers (a) and (b) were calculated.
 A: 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.
A: In a.) part, you get the probability as $\frac{81}{125}$. Therefore, in the long run, out of $125$ games, $A$ will win $81$ games and lose $44$ games. Therefore the odds in favor of A are $81:44$.
Similarly calculate for part b 
