$\begin{aligned} & \cos (\alpha+\beta)=\frac{4}{5} \Rightarrow \tan (\alpha+\beta)=\frac{3}{4} \\ & \sin (\alpha-\beta)=\frac{5}{13} \Rightarrow \tan (\alpha-\beta)=\frac{5}{12} \\ & \tan 2 \alpha=\tan \{(\alpha+\beta)+(\alpha-\beta)\}=\frac{\tan (\alpha+\beta)+\tan (\alpha-\beta)}{1-\tan (\alpha+\beta) \cdot \tan (\alpha-\beta)} \\ & =\frac{\frac{3}{4}+\frac{5}{12}}{1-\frac{3}{4} \times \frac{5}{12}}=\frac{56}{33}\end{aligned}$