3. Natural numbers $m, n$ satisfy $8 m+10 n>9 m n$. Then
$$
m^{2}+n^{2}-m^{2} n^{2}+m^{4}+n^{4}-m^{4} n^{4}=
$$
$\qquad$
3. 2 .

It is known that $\frac{8}{n}+\frac{10}{m}>9$. It is easy to see that one of the two natural numbers $m, n$ must be 1 (otherwise $\frac{8}{m}+\frac{10}{m} \leqslant \frac{8}{2}+\frac{10}{2}=9$ which is a contradiction). Without loss of generality, let $m=1$, then
$$
\begin{array}{l}
m^{2}+n^{2}-m^{2} n^{2}+m^{4}+n^{4}-m^{4} n^{4} \\
=1+n^{2}-n^{2}+1+n^{4}-n^{4}=2 .
\end{array}
$$