$z_{1}=2-i, z_{2}=-2+i$

$z_{1} z_{2}=(2-i)(-2+i)=-4+2 i+2 i-i^{2}=-4+4 i-(-1)=-3+4 i$

$\overline{z_{1}}=2+i$

$\therefore \frac{z_{1} z_{2}}{z_{1}}=\frac{-3+4 i}{2+i}$

On multiplying numerator and denominator by $(2-i),$ we obtain

$\frac{z_{1} z_{2}}{z_{1}}=\frac{(-3+4 i)(2-i)}{(2+i)(2-i)}=\frac{-6+3 i+8 i-4 i^{2}}{2^{2}+1^{2}}=\frac{-6+11 i-4(-1)}{2^{2}+1^{2}}$

$=\frac{-2+11 i}{5}=\frac{-2}{5}+\frac{11}{5} i$

On comparing real parts, we obtain

$\operatorname{Re}\left(\frac{z_{1} z_{2}}{\bar{z}_{1}}\right)=\frac{-2}{5}$