Solution:

(1)

$\frac{d y}{d x}=x+3 y \quad x_{0}=0, y_{0}=1, h=0.1$

$y_{1}^{(0)}=y_{0}+h . f\left(x_{0}, y_{0}\right)=1+0.1(3)=1.3$

$y_{1}^{n+1}=y_{0}+\frac{h}{2}\left[f\left(x_{0}, y_{0}\right)+f\left(x_{1}, y_{1}^{n}\right)\right]$

$\begin{array}{|c|c|c|c|c|}\hline \text { Iteration } & {x_{1}} & {y_{1}^{n}} & {x_{1} y_{1}^{n}} & {y_{1}^{n+1}} \\ \hline 0 & {0.1} & {1.3} & {4} & {1.35} \\ \hline 1 & {0.1} & {1.35} & {4.15} & {1.3575} \\ \hline 2 & {0.1} & {1.3575} & {4.1725} & {1.3587} \\ \hline\end{array}$

$y(0.1)=1.3587$

(2)

$x_{1}=0.1, y_{1}=1.3587$

$y_{2}^{0}=1.77631$

$\quad y_{2}^{n+1}=y_{1}+\frac{h}{2}\left[f\left(x_{2}, y_{2}\right)+f\left(x_{2}, y_{2}^{n}\right)\right]$

$\begin{array}{|c|c|c|c|c|c|}\hline \text { Iteration } & {x_{2}} & {y_{2}^{n}} & {x_{2} y_{2}^{n}} & {y_{2}^{n+1}} \\ \hline 0 & {0.2} & {1.77631} & {5.52893} & {1.8439} \\ \hline 1 & {0.2} & {1.8439} & {5.7317} & {1.8540} \\ \hline 2 & {0.2} & {1.8540} & {5.762} & {1.8556} \\ \hline\end{array}$

$y(0.2)=1.8556$