3. As shown in Figure 2, in the right triangular prism $A B C-A_{1} B_{1} C_{1}$, the base is an isosceles right triangle, $\angle A_{1} C_{1} B_{1}=90^{\circ}, A_{1} C_{1}=$ $1, A A_{1}=\sqrt{2}$, find $A B_{1}$, $A C_{1}$, and the plane angle between the plane $A A_{1} B_{1}$ and the plane $A B_{1} C_{1}$. (Express using inverse trigonometric functions)
3. $\operatorname{arctg} \sqrt{2}$.

Take the midpoint $D$ of $A_{1} B_{1}$, and connect $C_{1} D$. In the isosceles right $\triangle A_{1} B_{1} C_{1}$, since $C_{1} D$ is the median to the hypotenuse, we know that $C_{1} D \perp A_{1} B_{1}$. Also, by the properties of a right prism, $C_{1} D \perp A A_{1}$. Therefore, $C_{1} D \perp$ plane $A A_{1} B_{1} B$.

In the plane $A B_{1} C_{1}$, draw $C_{1} E \perp A B_{1}$, and connect $D E$. By the inverse of the three perpendiculars theorem, we get $D E \perp A B_{1}$. Thus, the required dihedral angle's plane angle is $\angle C_{1} E D$.

In the right $\triangle A A_{2} B_{1}$, given $A A_{1}=A_{1} B_{1}=\sqrt{2}$, and $D$ is the midpoint of $A_{1} B_{1}$, we can find
$$
\begin{array}{l}
D E=\frac{\sqrt{2}}{2} D B_{1} \\
=\frac{\sqrt{2}}{4} A_{1} B_{1}=\frac{1}{2} .
\end{array}
$$

In the isosceles right $\triangle A_{1} B_{1} C_{1}$, we can also find $C_{1} D=\frac{\sqrt{2}}{2}$.
In $\triangle C D E$, $\angle C D E=90^{\circ}$, we have $\operatorname{tg} \angle C_{1} E D=\frac{C_{1} D}{D E}=\sqrt{2}$.
$$
\therefore \angle C_{1} E D=\operatorname{arctg} \sqrt{2} \text {. }
$$

It can also be expressed as $\arcsin \frac{\sqrt{6}}{3}$ or $\arccos \frac{\sqrt{3}}{3}$.