how to find out coset representation of the following subgroup of the given group? $G$ is a group of $2\times 2$ matrices= $SL(2,\mathbb{Z})/\{I_2,-I_2\}$ where $SL(2,\mathbb{Z})$ is invertible matrix with entries in $\mathbb{Z}$ and determinant $1$.
then I know that G is generated by following two matrices
$$\begin{bmatrix}1&1\\0&1\end{bmatrix},\begin{bmatrix}0&1\\-1&0\end{bmatrix}$$
now Let $H$ be its subgroup:
$$H :=\{\begin{bmatrix}1+3s&3t\\3v&1+3w\end{bmatrix}\mid r,t,v,w \text{ are integers and  $det(A)=1$}\}$$
We can say $H$ is the subgroup of $G$ having matrices congruent to $I_2\text{ mod }3$.
further index of $H$ in $G$ is $12$. then I know coset reperesentatives of $H$ in $G$.
these are $$\begin{bmatrix}1&0\\0&1\end{bmatrix},\begin{bmatrix}1&-1\\0&1\end{bmatrix},\begin{bmatrix}0&1\\-1&0\end{bmatrix},\begin{bmatrix}1&1\\-1&0\end{bmatrix},\begin{bmatrix}1&-2\\0&1\end{bmatrix},\begin{bmatrix}0&1\\-1&1\end{bmatrix}\begin{bmatrix}2&1\\-1&0\end{bmatrix},\begin{bmatrix}1&0\\-1&1\end{bmatrix},\begin{bmatrix}0&1\\-1&2\end{bmatrix},\begin{bmatrix}1&0\\1&1\end{bmatrix},\begin{bmatrix}2&-1\\-1&1\end{bmatrix},\begin{bmatrix}1&-1\\-1&2\end{bmatrix}$$ 
how to find out this coset representation?

Well, I would do it the following way :
$$G\rightarrow PSL(2,\mathbb{F}_3) $$
$$A\mapsto A\text{ mod }3$$
Is clearly a group morphism whose kernel is $H$. Since $PSL(2,\mathbb{F}_3)$ is of cardinal $12$ and the coset representatives you found are different modulo $3$, it follows that it is surjective. Hence :
$$G/H=PSL(2,\mathbb{F}_3) $$
You may also go further : by making $PSL(2,\mathbb{F}_3)$  act on the $4$ lines of a $\mathbb{F}_3$-vector space of dimension $2$, you can show that the action of $PSL(2,\mathbb{F}_3)$ is faithfull hence $PSL(2,\mathbb{F}_3)$ can be identified to a subgroup of the symmetric group $S_4$ of cardinal $12$. Since there exists only one subgroup of $S_4$ of index $2$ which is the alternating group $A_4$ one realizes that :
$$G/H\text{ is isomorphic to } A_4 $$
