6. Given $\frac{\pi}{2}<\beta<\alpha<\frac{3 \pi}{4}, \cos (\alpha-\beta)=\frac{12}{13}, \sin (\alpha+\beta)=-\frac{3}{5}$, find the value of $\sin 2 \alpha$ $\qquad$ .
$\begin{array}{c}\text { 6. }-\frac{56}{65} . \\ \left.\left.\begin{array}{l}\alpha+\beta \in\left(\pi, \frac{3 \pi}{2}\right) \\ \sin (\alpha+\beta)=-\frac{3}{5}\end{array}\right\} \Rightarrow \alpha+\beta=\pi+\arcsin \frac{3}{5}, \quad \begin{array}{l}\alpha-\beta \in\left(0, \frac{\pi}{4}\right) \\ \cos (\alpha-\beta)=\frac{12}{13}\end{array}\right\} \Rightarrow \alpha-\beta=\arccos \frac{12}{13} . \\ \text { Therefore, } 2 \alpha=(\alpha+\beta)+(\alpha-\beta)=\pi+\arcsin \frac{3}{5}+\arccos \frac{12}{13}, \\ \sin 2 \alpha=-\sin \left[\arcsin \frac{3}{5}+\arccos \frac{12}{13}\right]=-\left(\frac{3}{5} \times \frac{12}{13}+\frac{4}{5} \times \frac{5}{13}\right)=-\frac{56}{65} .\end{array}$