(2) If $\tan \alpha=3 \tan \beta\left(0 \leqslant \beta \leqslant \alpha<\frac{\pi}{2}\right)$, then the maximum value of $\alpha-\beta$ is
(2) $\frac{\pi}{6}$ Hint: Let $x=\alpha-\beta$, then $0 \leqslant x<\frac{\pi}{2}$,
$$
\begin{aligned}
\tan x & =\tan (\alpha-\beta) \\
& =\frac{\tan \alpha-\tan \beta}{1+\tan \alpha \tan \beta}=\frac{2 \tan \beta}{1+3 \tan ^{2} \beta} \\
& =\frac{2}{\cot \beta+3 \tan \beta},
\end{aligned}
$$

When $\cot \beta=3 \tan \beta$, i.e., $\beta=\frac{\pi}{6}$, $\tan x$ reaches its maximum. At this time, $\alpha=\frac{\pi}{3}$, so the maximum value of $x=\alpha-\beta$ is $\frac{\pi}{6}$.