${\bf GL}(\color{blue}{2}, \color{red}{\mathbb R})$ 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 $GL(2, \mathbb R)$.
First of all you have to ask yourself if it is possibile that $H/K$ is a group, i.e if $K$ is normal in $H$.
In your case (please verify this fact) you have that $K$ is an abelian group (in particular is isomorphic to $(\mathbb{R},+,0)$ ) and is contained in the center of $H$ (I.e $AK=KA$ for each $A \in H$, and $K\in K$), so that $K$ is clearly normal in $H$.
Thus The quotient set $H/K=\{AK: A \in H\}$ inherits also a group structure.
Now we have to understand what is the multiplication in $H$ to compute easily the quotient set $H/K$.
Taking an element $A\in H$, I will denote $A$ as $A(a,b,c)$, where $A_{12}=a, A_{13}=b$ and $A_{23}=c$.
Then taking two matrices $A=A(a,b,c)$ and $B=B(a’,b’,c’)$, you have
$AB=AB(a+a’, b+ac’ +b’, c+c’)=BA$
and so if you have a general element $A=A(a,b,c)$, you can observe that $A’(a,0,c) $ is an element in its class on $H/K$, in fact
$A(a,b,c)=A’(a, 0, c)K(0, b+ac’+b’, 0) $
So that you can define the section map $s: H/K \to G$ Sending each class $AK$ to $A’(a,0,c)$, where $(G:=\{ A(a,0,c) : a,c \in \mathbb{R}\},+,0)$ is an additive group.
You can prove that this map is an isomorphism of groups and observing that $A(a,0,c)+B(a’,0,c’)=(A+B)(a+a’, 0, c+c’)$
then you can say that $H/K \cong (\mathbb{R}^2,+,0)$
Best Answer
$G\simeq \mathbb Z^4$ (the isomorphism is given by $\begin{bmatrix}a & b \\ c & d\end{bmatrix}\mapsto (a,b,c,d)$) and $H\simeq (2\mathbb Z)^4$ $\Rightarrow$ $G/H\simeq ({\mathbb Z}/2\mathbb Z)^4$ and this shows that $|G/H|=16$.