$$
\begin{aligned}
& \text { Let } a+b+c+d=x \text { and } e+f+g+h=y \\
& x^2+y^2=(x+y)^2-2 x y \\
& =(13+6+4+2-2-3-5-7)^2-2 x y \\
& =64-2 x y
\end{aligned}
$$
As we know $\mathrm{AM} \geq \mathrm{GM}$
$$
\begin{aligned}
& \frac{x+y}{2} \geq \sqrt{x y} \\
\Rightarrow \quad & \frac{8}{2} \geq \sqrt{x y} \\
\Rightarrow \quad & x y \leq 16
\end{aligned}
$$
$x^2+y^2$ is minimum when $x y$ is maximum.
The maximum value of $x y=16$.
The minimum value of $x^2+y^2=64-2 \times 16=32$
$\therefore$ The minimum values of
$$
(a+b+c+d)^2+(e+f+g+h)=32
$$