11. Let complex numbers $z_{1}, z_{2}$ satisfy $\operatorname{Re}\left(z_{1}\right)>0, \operatorname{Re}\left(z_{2}\right)>0$, and $\operatorname{Re}\left(z_{1}^{2}\right)=\operatorname{Re}\left(z_{2}^{2}\right)=2$ (where $\operatorname{Re}(z)$ denotes the real part of the complex number $z$).
(1) Find the minimum value of $\operatorname{Re}\left(z_{1} z_{2}\right)$;
(2) Find the minimum value of $\left|z_{1}+2\right|+\left|\overline{z_{2}}+2\right|-\left|\overline{z_{1}}-z_{2}\right|$.
(1) Let $z_{1}=a_{1}+b_{1} \mathrm{i}, z_{2}=a_{2}+b_{2} \mathrm{i}, a_{i}, b_{i} \in \mathbf{R}(i=1,2)$,

Then $z_{1}^{2}=a_{1}^{2}-b_{1}^{2}+2 a_{1} b_{1} \mathrm{i}, z_{2}^{2}=a_{2}^{2}-b_{2}^{2}+2 a_{2} b_{2} \mathrm{i} \Rightarrow a_{1}^{2}-b_{1}^{2}=a_{2}^{2}-b_{2}^{2}=2$.
Also, $z_{1} z_{2}=a_{1} a_{2}-b_{1} b_{2}+\left(a_{1} b_{2}-a_{2} b_{1}\right) \mathrm{i} \Rightarrow \boldsymbol{\operatorname { R e }}\left(z_{1} z_{2}\right)=a_{1} a_{2}-b_{1} b_{2}$,
$$
\begin{array}{l}
\left(a_{1} a_{2}-b_{1} b_{2}\right)^{2}-\left(a_{1}^{2}-b_{1}^{2}\right)\left(a_{2}^{2}-b_{2}^{2}\right)=a_{1}^{2} b_{2}^{2}+a_{2}^{2} b_{1}^{2}-2 a_{1} a_{2} b_{1} b_{2} \\
=\left(a_{1} b_{2}-a_{2} b_{1}\right)^{2} \geqslant 0 \Rightarrow\left(a_{1} a_{2}-b_{1} b_{2}\right)^{2} \geqslant\left(a_{1}^{2}-b_{1}^{2}\right)\left(a_{2}^{2}-b_{2}^{2}\right)=4 .
\end{array}
$$

Since $a_{1}, a_{2}>0 \Rightarrow a_{1} a_{2}-b_{1} b_{2}>0 \Rightarrow a_{1} a_{2}-b_{1} b_{2} \geqslant 2$, equality holds when $a_{1} b_{2}=a_{2} b_{1}$.
Therefore, the minimum value of $\boldsymbol{\operatorname { R e }}\left(z_{1} z_{2}\right)$ is 2.
(2) As shown in the figure, since $z_{1}, z_{2}$ correspond to points
$A, B$ on the right branch of the hyperbola $x^{2}-y^{2}=2$,
where $F_{1}(-2,0), F_{2}(2,0)$. Then
$$
\begin{array}{l}
\left|z_{1}+2\right|+\left|\overline{z_{2}}+2\right|-\left|\overline{z_{1}}-z_{2}\right| \\
=\left|A F_{1}\right|+\left|B^{\prime} F_{1}\right|-\left|A^{\prime} B\right| \\
=2 \sqrt{2}+\left|A F_{2}\right|+2 \sqrt{2}+\left|B^{\prime} F_{2}\right|-\left|A B^{\prime}\right| \\
=4 \sqrt{2}+\left|A F_{2}\right|+\left|B^{\prime} F_{2}\right|-\left|A B^{\prime}\right| \geqslant 4 \sqrt{2},
\end{array}
$$
$=4 \sqrt{2}+\left|A F_{2}\right|+\left|B^{\prime} F_{2}\right|-\left|A B^{\prime}\right| \geqslant 4 \sqrt{2}$, equality holds when $A, F_{2}, B^{\prime}$ are collinear.
Therefore, the minimum value of $\left|z_{1}+2\right|+\left|\overline{z_{2}}+2\right|-\left|\overline{z_{1}}-z_{2}\right|$ is $4 \sqrt{2}$.