Let $w=z+2.$ Then given: $|w|=1$ and rac{z+1}{z+2}=rac{(w-2)+1}{w}=rac{w-1}{w}=1-rac{1}{w}.

Since $|w|=1$, we have $\frac{1}{w}=\overline{w}.$ Therefore, $\frac{z+1}{z+2}=1-\overline{w}.$

Let $w=x+iy,$ where $x,y\in\mathbb R$ and $x^2+y^2=1.$ Then $\overline{w}=x-iy,$ so $1-\overline{w}=1-x+iy.$ Hence, $\operatorname{Im}\left(\frac{z+1}{z+2}\right)=y.$

Given $\operatorname{Im}\left(\frac{z+1}{z+2}\right)=\frac15,$ we get $y=\frac15.$

Using $x^2+y^2=1$: $x^2+\left(\frac15\right)^2=1$ $x^2=1-\frac1{25}=\frac{24}{25}$ $|x|=\frac{2\sqrt6}{5}.$

We need $\left|\operatorname{Re}(\overline{z+2})\right|=|\operatorname{Re}(\overline{w})|.$ Since $w=x+iy$, we have $\overline{w}=x-iy,$ so $\operatorname{Re}(\overline{w})=x.$ Therefore, $\left|\operatorname{Re}(\overline{z+2})\right|=|x|=\frac{2\sqrt6}{5}.$

So the correct option is $\boxed{\text{A } \frac{2\sqrt6}{5}}.$