Solution
$|\mathrm{z}+2|=1$ Let $\mathrm{z}+2=\cos \theta+\mathrm{i} \sin \theta \Rightarrow \overline{\mathrm{z}+2}=\frac{1}{\mathrm{z}+2}=\cos \theta-\mathrm{i} \sin \theta$
\[
\begin{aligned}
& \Rightarrow \frac{z+1}{z+2}=1-\frac{1}{z+2}=1-(\cos \theta-i \sin \theta)=(1-\cos \theta)-i \sin \theta \\
& \operatorname{Im}\left(\frac{z+1}{z+2}\right)=\sin \theta=\frac{1}{5} \Rightarrow \cos \theta= \pm \sqrt{1-\frac{1}{25}}= \pm \frac{2 \sqrt{6}}{5} \Rightarrow|\operatorname{Re}(\overline{z+2})|=|\cos \theta|=\frac{2 \sqrt{6}}{5}
\end{aligned}
\]
.
