If

$\frac{(1+i)^{2}}{2-i}=\frac{1^{2}+i^{2}+2 i}{2-i}$

$=\frac{1-1+2 i}{2-i} \quad\left[\because i^{2}=-1\right]$

$=\frac{2 i}{2-1} \times \frac{2+i}{2+i}$

$=\frac{2 i(2+i)}{2^{2}-i^{2}}$

$=\frac{4 i+2 i^{2}}{4+1} \quad \because i^{2}=-1 \mid$

$=\frac{4 i-2}{5}$

$=\frac{-2}{5}+\frac{4}{5} i$   ....(1)

It is given that,

$\frac{(1+i)^{2}}{2-i}=x+i y$

$\Rightarrow-\frac{2}{5}+\frac{4}{5} i=x+i y$     $[\operatorname{From}(1)]$

$\Rightarrow x=-\frac{2}{5}$ and $y=\frac{4}{5}$

$\therefore x+y=\frac{-2}{5}+\frac{4}{5}$

$=\frac{2}{5}$

Thus, $x+y=\frac{2}{5}$.