1. Calculate $\tan 22.5^{\circ}=$
2. $\sqrt{2}-1$.
As shown in Figure 4, it is easy to see that
$$
\begin{array}{l}
\angle B A C=45^{\circ}, \\
\angle C=90^{\circ}, \\
A C=B C=1, \\
A B=\sqrt{2} .
\end{array}
$$

Construct the angle bisector $A D$ of $\angle B A C$. Then
$$
\begin{array}{c}
\angle D A C=22.5^{\circ} . \\
\text { In right } \triangle A C D, \\
\tan 22.5^{\circ}=\frac{D C}{A C}=\frac{D C}{1}=D C .
\end{array}
$$

By the angle bisector property of a triangle, we have
$$
\begin{array}{l}
\frac{B D}{D C}=\frac{A B}{A C} \Rightarrow \frac{1-D C}{D C}=\frac{\sqrt{2}}{1} \\
\Rightarrow D C=\sqrt{2}-1 .
\end{array}
$$

Therefore, $\tan 22.5^{\circ}=\sqrt{2}-1$.