7. Let $\sin \alpha+\sin \beta=\frac{4}{5} \sqrt{2}, \cos \alpha+\cos \beta=\frac{4}{5} \sqrt{3}$, then $\tan \alpha+\tan \beta=$
According to the problem, $\left\{\begin{array}{l}2 \sin \frac{\alpha+\beta}{2} \cos \frac{\alpha-\beta}{2}=\frac{4}{5} \sqrt{2}, \\ 2 \cos \frac{\alpha+\beta}{2} \cos \frac{\alpha-\beta}{2}=\frac{4}{5} \sqrt{3}\end{array} \Rightarrow \tan \frac{\alpha+\beta}{2}=\frac{\sqrt{6}}{3}\right.$,
thus $\sin (\alpha+\beta)=\frac{2 \times \frac{\sqrt{6}}{3}}{1+\frac{2}{3}}=\frac{2 \sqrt{6}}{5}, \cos (\alpha+\beta)=\frac{1-\frac{2}{3}}{1+\frac{2}{3}}=\frac{1}{5}$.
Also, $2+2 \cos (\alpha-\beta)=\frac{32}{25}+\frac{48}{25}=\frac{16}{5} \Rightarrow \cos (\alpha-\beta)=\frac{3}{5}$,

so $\tan \alpha+\tan \beta=\frac{\sin (\alpha+\beta)}{\cos \alpha \cos \beta}=\sqrt{6}$.