Let $G = \langle x,y \mid x^2,y^3 \rangle \cong {\rm PSL}(2,\mathbb{Z})$.
Question 1. Yes. Let $H < G$, and consider the permutation action of $G$ on the (left or right) cosets of $H$ in $G$. If $|G:H| < 6$, then it is not possible for the images of both $x$ and $y$ to act fixed point freely, and so $H$ contains a conjugate of $x$ or $y$ and hence cannot be free.
Question 2. No, but the two subgroups that you have found are the only two normal subgroups of index $6$ in $G$. You can see this by observing that there is essentially only one surjective group homomorphism from $G$ to each of $C_6$ and $S_3$ (i.e. up to equivalence under an automorphism of $C_6$ or $S_3$), so there are only two possible kernels $H$.
By a computer calculation, I found that there is also one conjugacy class of non-normal subgroups $H$ with $|G:H| = 6$ and with $H$ free of rank $2$, and a representative of this class is $H=\langle yx, y^{-1}(xy)^3 \rangle$. The quotient of $G$ by the core of $H$ is isomorphic to $S_4$, and there are three conjugates of $H$ in $G$.
Question 3. I can only say here that the reason is that we have a proof that this is the case!
Note that the Kurosh Subgroup Theorem says that any subgroup of $G$ is a free product of conjugates of $\langle x \rangle$, $\langle y \rangle$ and a free subgroup of $G$. So, for $H \lhd G$, if $H$ does not contain $x$ or $y$, then it must be free. I believe that the rank of free subgroups of free products can be calculated using Euler Characteristics, but I don't know the details.
Best Answer
$\left( \begin{array}{cc}1&1\\0&1\end{array} \right)^n = \left( \begin{array}{cc}1&n\\0&1\end{array} \right)$ does not commute with $\left( \begin{array}{cc}1&0\\1&1\end{array} \right)^n = \left( \begin{array}{cc}1&0\\n&1\end{array} \right)$ for any $n > 0$, so ${\rm PSL}(2,{\mathbb Z})$ has no abelian subgroup of finite index.