jmc

Note that the sum of the elements in the set is 8. Let $x=a+b+c+d$, so $e+f+g+h=8-x$. Then

$$\begin{array}{rl}(a+b+c+d{)}^2& +(e+f+g+h{)}^2=x^2+(8-x{)}^2\\ & =2x^2-16x+64=2(x-4{)}^2+32\ge 32.\end{array}$$The value of 32 can be attained if and only if $x=4$. However, it may be assumed without loss of generality that $a=13$, and no choice of $b,c$, and $d$ gives a total of 4 for $x$. Thus $(x-4{)}^2\ge 1$, and $$(a+b+c+d{)}^2+(e+f+g+h{)}^2=2(x-4{)}^2+32\ge \boxed{34}.$$A total of 34 can be attained by letting $a,b,c$, and $d$ be distinct elements in the set $\{-7,-5,2,13\}$.