$$\begin{aligned} & |z+2|=1 \\ & \operatorname{Im}_m\left(\frac{z+1}{z+2}\right)=\frac{1}{5} \\ & |\operatorname{Re}(\overline{z+2})|=? \end{aligned}$$

Let $$z=x+i y$$

$$\begin{aligned} & \because|z+2|=1 \Rightarrow(x+2)^2+y^2=1 \quad \ldots(1) \\ & I_m\left(\frac{z+1}{z+2}\right)=\frac{1}{5} \Rightarrow I_m\left(\frac{x+i y+1}{x+i y+2}\right)=\frac{1}{5} \\ & \Rightarrow I_m\left|\frac{[(x+1)+i y][(x+)-i y]}{(x+2)^2+y^2}\right|=\frac{1}{5} \end{aligned}$$

$$\frac{y(x+2)-y(x+)}{(x+2)^2+y^2}=\frac{1}{5}\quad \text{.... (2)}$$

$$\Rightarrow y=\frac{1}{5}$$

Substituting in equation (1)

$$\begin{aligned} & (x+2)^2+\frac{1}{25}=1 \\ & (x+2)^2=\frac{24}{25} \\ & \Rightarrow x=-2 \pm \frac{\sqrt{24}}{5} \\ & |\operatorname{Re}(\overline{x+i y+2})| \\ & =x+2= \pm \frac{\sqrt{24}}{5}=\frac{2 \sqrt{6}}{5} \end{aligned}$$