(i)
$\begin{aligned}
&\left(\frac{z_1 z_2}{\bar{z}_1}\right)=\frac{(2-i)(-2+i)}{2-i}=\frac{-(2-i)^2}{2+i} \\
&=\frac{-\left(4+i^2-4 i\right)}{2+i}=\frac{-(3-4 i)}{2+i}=\frac{-3+4 i}{2+i} \\
&=\frac{-3+4 i}{2+i} \times \frac{2-i}{2-i}=\frac{-6+3 i+8 i-4 i^2}{4-i^2} \\
&=\frac{-2+11 i}{5}=\frac{-2}{5}+\frac{11}{5} i \\
\therefore \quad & \operatorname{Re}\left(\frac{z_1 z_2}{\bar{z}_1}\right)=\frac{-2}{5}
\end{aligned}$
(ii) $\frac{1}{z_1 \bar{z}_1}=\frac{1}{(2-i)(2+i)}=\frac{1}{4-i^2}=\frac{1}{5}=\frac{1}{5}+i .0$
$\therefore \quad \operatorname{Im}\left(\frac{1}{z_1 \bar{z}_2}\right)=0$