5. (2004 National College Entrance Examination) Given in acute triangle $\triangle ABC$, $\sin (A+B)=\frac{3}{5}, \sin (A-B)=\frac{1}{5}$.
(1) Prove: $\tan A=2 \tan B$;
(2) Let $AB=3$, find the altitude from $C$ to $AB$.
5. (1)
$$
\left.\begin{array}{l}
\sin (A+B)=\sin A \cos B+\cos A \sin B=\frac{3}{5} \\
\sin (A-B)=\sin A \cos B-\cos A \sin B=\frac{1}{5}
\end{array}\right\} \frac{\tan A}{\tan B}=\frac{\sin A \cos B}{\cos A \sin B}=2
$$

So $\tan A=2 \tan B$.
(2) Since $\triangle ABC$ is an acute triangle, draw $CD \perp AB$ at $D$. It is easy to know: $AD=1, BD=2$, then $AC^{2}=CD^{2}+AD^{2}=CD^{2}+1 ; BC^{2}=CD^{2}+4$
From $\sin (A+B)=\frac{3}{5}$ we get $\sin C=\frac{3}{5} \quad$ so $\quad \cos C=\frac{4}{5}$
In $\triangle ABC$ by the cosine rule we get: $AB^{2}=AC^{2}+BC^{2}-2 AB \cdot BC \cos C$ that is:
$$
9=CD^{2}+1+CD^{2}+4-\frac{8}{5} \sqrt{1+CD^{2}} \cdot \sqrt{4+CD^{2}}
$$

Solving this, we get: $CD=2+\sqrt{6}$