Probabilities in choosing committees

Your calculation for the first question is wrong: there are just $\binom{10}3$ ways to choose a committee consisting of Ryan and three girls, since the only freedom of choice is in the selection of the girls. The probability is therefore

$$\frac{\binom{10}3}{\binom{22}4}=\frac{120}{7315}\approx0.0164\;,$$

or $0.02$ when rounded to two decimal places. It does, therefore, appear that the official answer slipped the decimal point.

The second question asks for a conditional probability: given that Ryan is on the committee, what is the probability that the other three committee members are girls. There are $\binom{21}3$ committees that include Ryan as one of their members, and as we saw in the first problem, $\binom{10}3$ of these have three girls. The probability is therefore

$$\frac{\binom{10}3}{\binom{21}3}=\frac{120}{1330}\approx0.09\;,$$

and in this case the official answer is correct.

The first problem is similar to a problem you already asked. There are $\binom{22}{4}$ possible committees, by assumption all equally likely.

Now we ask how many committees there are that consist of Ryan and $3$ girls. There are $\binom{10}{3}$ ways of choosing these girls.

The second problem is more subtle. It is not clear at what stage you are. In particular, it is not clear whether the machinery of conditional probability has already been introduced. We will assume that it has not.

Your analysis is correct. Given that Ryan has to be on the committee, we are choosing $3$ people from $22$.