For the rest of this post, let
$$ A=\begin{pmatrix} 1 & 0\\ 2 & 1\end{pmatrix}, B=\begin{pmatrix} 1 & 2\\ 0 & 1\end{pmatrix}, C=\begin{pmatrix} -1 & 0\\ 0 & -1\end{pmatrix}, D=\begin{pmatrix} -1 & 0\\ 0 & 1\end{pmatrix}.$$
Note that all of $A$, $B$, $C$ and $D$ live in $\Gamma(2)$.
First, let's consider the case of $G=SL(2,\mathbb{Z}$), which is the case I think you wanted (so in your notation, we have the requirement that $ad-bc=1$). Note that in this case, $D\notin G$.
Proposition: $A$, $B$, and $C$ generate $\Gamma(2)$.
Proof: Define a mapping from $f:\ \Gamma(2)\rightarrow \mathbb{Z}^+$ by the formula
$$ f:\ \begin{pmatrix} a & b\\ c & d\end{pmatrix}\mapsto |a|+|c|.$$
Let $\mathfrak{H}$ be the subgroup of $\Gamma(2)$ generated by $A$, $B$, and $C$, and let $X$ be an arbitrary element of $\Gamma(2)$. We will be done if we can show $X\in \mathfrak{H}$.
To this end, pick an element $Y\in \mathfrak{H}X$ [the right coset of $\mathfrak{H}$ containing $X$] for which $f(Y)$ is minimal.
Now letting $Y=\begin{pmatrix} a & b\\ c & d\end{pmatrix}$, consider the following cases:
- $c=0$.
We know $ad-bc=1$, and so in this case $a=d=\pm 1$. But then $Y$ (or $YC$) must be a power of $B$, since
$$ B^n=\begin{pmatrix} 1 & 2n\\ 0 & 1\end{pmatrix}.$$
This means $Y\in\mathfrak{H}\cap\mathfrak{H}X$, so that $\mathfrak{H}=\mathfrak{H}X$, or $X\in \mathfrak{H}$.
- $c\neq0$, and $|a| > |c|$.
Then there exists an $n\in\mathbb{Z}$ such that $-|c| < a+2nc < |c|$ [strict inequality because $a$ is odd and $c$ is even], and then
$$ B^nY=\begin{pmatrix} 1 & 2n\\ 0 & 1\end{pmatrix}\begin{pmatrix} a & b\\ c & d\end{pmatrix}=\begin{pmatrix} a+2nc & b+2nd\\ c & d\end{pmatrix},$$
so that $f(B^nY)=|a+2nc| + |c| < |c| + |c| < |a| + |c|$, contradicting the choice of $Y$. In other words, this case does not happen.
- $c\neq 0$, and $|a| < |c|$. Then a similar argument to the one above, using $A$ instead of $B$, leads also to a contradiction. The proof is now complete.
Thus we see that if $Z=\langle C\rangle$, then $\Gamma(2)/Z$ is generated by $A$ and $B$. But the group generated by $A$ and $B$ is free, by the Ping-Pong Lemma.
I think this is enough for the question, but if you actually meant $G=GL(2,\mathbb{Z})$, one can still say a lot. In this case, $\Gamma(2)/Z$ is not free, but has $F_2$ as a subgroup of index 2. There are only a handful of groups which possess $F_2$ as a subgroup of index 2, and in this case one gets the isomorphism $\Gamma(2)/Z\cong F_2\rtimes C_2$, where $F_2=\langle A, B\rangle$ and $C_2=\langle D\rangle$, with $D$ acting via $A^D=A^{-1}, B^D=B^{-1}$.
The name of the group is $\text{GL}_2(\mathbf{Z})$.
Indeed, note that
$$
\left(
\begin{smallmatrix}
1&0\\
1&1
\end{smallmatrix}
\right)=
\left(
\begin{smallmatrix}
0&1\\
1&0
\end{smallmatrix}
\right)
\left(
\begin{smallmatrix}
1&1\\
0&1
\end{smallmatrix}
\right)
\left(
\begin{smallmatrix}
0&1\\
1&0
\end{smallmatrix}
\right),$$
and thus your group contains $\text{SL}_2(\mathbf{Z})$ (see, e.g., Corollary 2.6 in http://www.math.uconn.edu/~kconrad/blurbs/grouptheory/SL%282,Z%29.pdf). On the other hand, your group is not contained in $\text{SL}_2(\mathbf{Z})$, and since $[\text{GL}_2(\mathbf{Z}):\text{SL}_2(\mathbf{Z})]=2$ your group must be $\text{GL}_2(\mathbf{Z})$.
Best Answer
${\bf GL}(\color{blue}{2}, \color{red}{\mathbb R})$ is called the $\bf{G}$eneral $\bf L$inear group consisting of $\color{blue}{2} \times \color{blue}2$ invertible matrices with $\color{red}{\text{real}}$ valued entries.
The group operation is matrix multiplication.
$\begin{pmatrix} 1 & 0 \\ 0 & 1\end{pmatrix}$ IS included in $GL(2\mathbb R)$. Indeed, it is the identity element in $GL(2 \mathbb R)$.
In (a), you are being asked to show that the set of all $2\times 2$ matrices whose determinant is $1$ is a subgroup of $GL(2, \mathbb R)$.
In (b), you are being asked to show that all $2\times 2$ diagonal matrices (whose entries on the diagonal are non-zero) is a subgroup of $GL(2, \mathbb R)$.