(3) In $\triangle A B C$, the lengths of the sides opposite to angles $A, B, C$ are $a, b, c$ respectively, and $S_{\triangle}=a^{2}-$ $(b-c)^{2}$. Then $\tan \frac{A}{2}=$ $\qquad$.
(3) $\frac{1}{4}$ Hint: Since
$$
\begin{aligned}
a^{2}-(b-c)^{2} & =a^{2}-\left(b^{2}-2 b c+c^{2}\right) \\
& =-\left(b^{2}+c^{2}-a^{2}\right)+2 b c \\
& =-2 b c \cos A+2 b c,
\end{aligned}
$$

we get
$$
\frac{1}{2} b c \sin A=-2 b c \cos A+2 b c,
$$

so
$$
4-4 \cos A=\sin A \text {, i.e., } 8 \sin ^{2} \frac{A}{2}=2 \sin \frac{A}{2} \cos \frac{A}{2} \text {, }
$$

thus $\tan \frac{A}{2}=\frac{1}{4}$.