1. Given $a+b=8, a^{2} b^{2}=4$. Then $\frac{a^{2}+b^{2}}{2}-a b=$ $\qquad$ .
$$
\begin{array}{l}
\text { Given } a+b=8, a^{2} b^{2}=4 \Rightarrow a b= \pm 2 \\
\Rightarrow \frac{a^{2}+b^{2}}{2}-a b=\frac{(a+b)^{2}-4 a b}{2} \\
\quad=\frac{8^{2} \pm 8}{2}=28 \text { or } 36 .
\end{array}
$$