Solution:

(1)

$\frac{d y}{d x}=\frac{y-x}{y+x} \quad x_{0}=0, y_{0}=1, h=0.2$

$f(x, y)=\frac{y-x}{y+x}$

$k_{1}=h . f\left(x_{0}, y_{0}\right)=0.2 f(0,1)=0.2$

$k_{2}=h . f\left(x_{0}+\frac{h}{2}, y_{0}+\frac{k_{1}}{2}\right)=0.2 . f(0.1,1.1)=0.1666$

$k_{3}=h . f\left(x_{0}+\frac{h}{2}, y_{0}+\frac{k_{2}}{2}\right)=0.2 \cdot f(0.1,1.0833)=0.1661$

$k_{4}=h . f\left(x_{0}+h, y_{0}+k_{3}\right)=0.2 f(0.2,1.1661)=0.1414$

$k=\frac{k_{1}+2 k_{2}+2 k_{3}+k_{4}}{6}=\frac{0.2+2(0.1666)+2(0.1661)+0.1414}{6}=0.1678$

$\therefore y(0.2)=y_{0}+k=1+0.1678=1.1678$

(2)

$\begin{aligned} x_{1} &=0.2, y_{2}=1.1678, h=0.2 \\ k_{5} &=h . f\left(x_{1}, y_{1}\right)=0.2 f(0.2,1.1678)=0.1415 \end{aligned}$

$k_{6}=h . f\left(x_{1}+\frac{h}{2}, y_{1}+\frac{k_{5}}{2}\right)=0.2 . f(0.3,1.23855)=0.1220$

$k_{7}=h . f\left(x_{1}+\frac{h}{2}, y_{1}+\frac{k_{6}}{2}\right)=0.2 \cdot f(0.3,1.2285)=0.1214$

$k_{8}=h . f\left(x_{1}+h, y_{1}+k_{7}\right)=0.2 f(0.4,1.2892)=0.1052$

$k *=\frac{k_{5}+2 k_{6}+2 k_{7}+k_{8}}{6}=\frac{0.1415+2(0.1220)+2(0.1215)+0.1052}{6}=0.1222$

$y(0.4)=y_{1}+k *=1.1678+0.1222=1.290$