3. As shown in Figure 1, given a square $A B C D$ with a side length of $5, E$ and $F$ are two points outside the square, $B E=D F=4, A E=C F=$ 3. Then $E F=$ . $\qquad$
3. $7 \sqrt{2}$.

It is known that, $\angle A E B=\angle C F D=90^{\circ}$.
Extend $E A$ and $F D$ to intersect at point $G$.
Since $\angle A D G=90^{\circ}-\angle C D F=\angle D C F=\angle B A E$, $\angle D A G=90^{\circ}-\angle B A E=\angle A B E$,
we know that $\triangle A G D \cong \triangle B E A$.
Therefore, $E G=F G=7$.
Thus, $E F=7 \sqrt{2}$.