Not only these subgroups are finitely presented, they are all finite free products of cyclic groups; most of them (for sufficiently large $n$) are actually free of finite rank (once congruence subgroup contains no elements of order 2 and 3). For instance, you can easily check that $\Gamma(n)$ is torsion-free for all $n\ge 2$ by looking at traces for $n\ge 3$ (since $1\ne 2$ mod $n\ge 2$ and $2\ne 0$ mod $n\ge 3$) and by looking at matrix coefficients for $n=2$. Rank is easily computable if you know index of the congruence subgroup in the modular group. The magic formula is multiplicativity of the Euler characteristic: For the modular group $\Gamma$, $\chi=-1+\frac{1}{2} + \frac{1}{3}=-\frac{1}{6}$. If $\Gamma'\subset \Gamma$ is a subgroup of index $i$ then $\chi(\Gamma')=i \chi(\Gamma)$. If $\Gamma$ is free of rank $r$ then $\chi(\Gamma)= 1-r$. For instance, to find index $i$ for $\Gamma(n)$, compute the order of the quotient group $SL(2, Z_n)/\pm I$. There is a closed formula for the order of this group (in terms of prime factors of $n$) which will tell you what the index is:
If $n$ is the product of powers of primes
$\prod_i p_i^{k_i}$ then
$$
|PSL(2,Z_n)|= \frac{n^3}{2} ~~~\prod_i (1- p_i^{-2}).
$$
