3. In rectangle $A B C D$, $A B=4, A D=3$, a ray of light is emitted from the midpoint $P_{0}$ of $A B$ to point $P_{1}$ on side $B C$, reflects to point $P_{2}$ on side $C D$, then reflects to point $P_{3}$ on side $D A$, and finally reflects to point $P_{4}$ on side $A B$. If $P_{0} P_{4}=1$, then $\tan \angle B P_{0} P_{1}=$ $\qquad$
3. $\frac{6}{7}$ or $\frac{2}{3}$.

When point $P_{4}$ is to the left of point $P_{0}$, by symmetry we get Figure 7.
Since $P_{0} P_{4}=1, P_{0}$ is the midpoint of $A B$, and $A B=4, A D=3$, then $A P_{4}=1, E A=1, E F=6, F G=10, P_{0} H=7$. Therefore, $\tan \angle B P_{0} P_{1}=\frac{6}{7}$.
Similarly, when point $P_{4}$ is to the right of point $P_{0}$, we have $\tan \angle B P_{0} P_{I}=\frac{2}{3}$.