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.
It's of infinite index and generates $\mathrm{PSL}_2(\mathbf{Z})$ as normal subgroup (hence is not normal).
Let me first check that it generates $\mathrm{PSL}_2(\mathbf{Z})$ as normal subgroup. Write $U=TS$. (All equalities are meant in $\mathrm{PSL}_2(\mathbf{Z})$.) So $U^3=1$. Modding out $\langle S,U\rangle$ by the relators $S,(US^{-1})^2$ is equivalent to mod out by $S,U^2$, and hence to mod out by $S,U$ since $U^3=1$. Hence the quotient is trivial, which is precisely the given assertion.
Next, let me check that it has infinite index. I use that $\mathrm{PSL}_2(\mathbf{Z})$ is the free product of $\langle S\rangle$ and $\langle U\rangle$.
Every free product $A\ast B$ can be written as $B^{\ast A}\rtimes A$, where the free factors in the kernel are the $aBa^{-1}$ when $a$ ranges over $A$. Here we thus write $\mathrm{PSL}_2(\mathbf{Z})= (\langle U\rangle\ast \langle SUS^{-1}\rangle)\rtimes \langle S\rangle$. Write $V=SUS^{-1}$: then $T^2=(US^{-1})^2=US^{-1}US=UV$ and $SUS^{-1}=VU$. Thus, in the free subgroup $F$ of index two $\langle U,V\rangle$, the intersection with $G$ is generated by $UV,VU$.
Let us check that it has infinite index. Indeed, inside $F$, modding out by $UV,VU$ yields an infinite cyclic group. So the normal subgroup of $F$ generated by $G\cap F$ has infinite index, and hence $G\cap F$ has infinite index in $F$, which in turn implies that $G$ has infinite index in $\mathrm{PSL}_2(\mathbf{Z})$.
Maybe you can check what this strategy provides when replacing $T^2$ with $T^k$.
Best Answer
Yes, your choice for $S$ and $T$ generate the modular group $\Gamma=PSL_2(\mathbb{Z})$. Other choices would also work.
The standard choice of generators is $$ \Gamma = \left \langle \left[ \begin{matrix} 1 & 1 \\ 0 & 1 \end{matrix} \right], \left[ \begin{matrix} 0 & 1 \\ -1 & 0 \end{matrix} \right] \right \rangle.$$
You already have $T=\left[ \begin{matrix} 1 & 1 \\ 0 & 1 \end{matrix} \right]$. Notice that $TST=\left[ \begin{matrix} 0 & -1 \\ 1 & 0 \end{matrix} \right] = - \left[ \begin{matrix} 0 & 1 \\ -1 & 0 \end{matrix} \right]$. Since we can obtain the standard generators for $\Gamma$ as products of $S$, $T$, and their inverses, we conclude that $S$ and $T$ generate $\Gamma$.
Just to clarify for anyone not familiar with all the matrix groups, $PSL_2(\mathbb{Z})=SL_2(\mathbb{Z})/\left\{1,-1\right\}$ is the group of 2x2 matrices over the integers with determinant 1, modulo the subgroup of scalar transformations, which in this case is just $\{1,-1\}$. That's why it doesn't matter if I obtain $\left[ \begin{smallmatrix} 0 & 1 \\ -1 & 0 \end{smallmatrix} \right]$ or $\left[ \begin{smallmatrix} 0 & -1 \\ 1 & 0 \end{smallmatrix} \right]$; they represent the same element in $PSL_2(\mathbb{Z})$.