\noindent{\bf Ans}: 

Let 

\begin{eqnarray*}
&&g_1({\bf x}) = \frac{1}{5}\left( 0.02 - \sin\left( \frac{6y}{1+y^2} \right) \right)\\
&&g_2({\bf x}) = \frac{1}{2}(0.01-\sin(x))(1+x)
\end{eqnarray*}
and
\begin{eqnarray*}
&&{\bf G}({\bf x}) = (g_1({\bf x}),\;g_2({\bf x}))^t\\
&&\bar{{\bf G}}({\bf x}) = \alpha {\bf x} + (I-\alpha){\bf G}({\bf x})
\end{eqnarray*}
where $\alpha$ is a $2 \times 2$ matrix and $I$ is the identity matrix.
One could choose 
$$
\alpha = (D{\bf G}({\bf x}_0)-I)^{-1}D{\bf G}({\bf x}_0).
$$
{\bf(up to here = 5 pts)}

If ${\bf x}_0 = (0,\;0)^t$, then the iteration gives 
$$
{\bf x}_* \approx (-4.882452175e-03,\;7.404885005e-03)^t.\;\textrm{{\bf(10 pts)}}
$$