4. As shown in Figure 1, in the right trapezoid $A B C D$, $A D / / B C$, $C D \perp A D, E$ is a point on $C D$, $\angle A B E=45^{\circ}, B C=C D$ $=6, A E=5$. Then the value of $\tan \angle A E B$ is ( ).
(A) 1.5
(B) 2
(C) 3
(D) None of the above
4.D.

As shown in Figure 6, draw $B F \perp A D$ at point $F$, extend $A F$ to point $G$ such that $F G=C E$, and connect $B G$.
Therefore, $\triangle B F G \cong \triangle B C E$.
We have $\angle A B G=\angle A B E=45^{\circ}, B E=B G$.
Also, $A B=A B$, so $\triangle A B G \cong \triangle A B E$, which gives
$$
A G=A E=5, \angle G=\angle A E B \text {. }
$$

Let $G F=x$, then
$$
C E=x, D E=6-x, A F=5-x, A D=1+x \text {. }
$$

In the right triangle $\triangle A D E$, by the Pythagorean theorem we have
$$
A D^{2}+D E^{2}=A E^{2} \text {, }
$$

which is $(6-x)^{2}+(1+x)^{2}=5^{2}$.
Solving this, we get $x=2$ or 3.
Thus, $\tan \angle A E B=\tan \angle B G F=\frac{B F}{G F}=\frac{6}{x}=3$ or 2.