6. 已知 $\alpha$ 为第三象限角, $\cos 2 \alpha=-\frac{3}{5}$, 则 $\tan \left(\frac{\pi}{4}+2 \alpha\right)=$
意 $\sin 2 \alpha=\frac{4}{5}, \therefore \tan 2 \alpha=-\frac{4}{3}$.

$\therefore \tan \left(\frac{\pi}{4}+2 \alpha\right)=\frac{\tan \frac{\pi}{4}+\tan 2 \alpha}{1-\tan \frac{\pi}{4} \tan 2 \alpha}=\frac{1-\frac{4}{3}}{1+\frac{4}{3}}=-\frac{1}{7}$.