7. Given $\alpha, \beta \in\left(0, \frac{\pi}{2}\right), \cos \alpha=\frac{4}{5}, \tan (\alpha-\beta)=-\frac{1}{3}$, find $\cos \beta$.


$\begin{array}{l}\text { 7. } \frac{9 \sqrt{10}}{50} \text {. } \\ \cos =\frac{4}{5}, \alpha \in\left(0, \frac{\pi}{2}\right) \Rightarrow \alpha=\arccos \frac{4}{5}, \alpha, \beta \in\left(0, \frac{\pi}{2}\right), \tan (\alpha-\beta)=-\frac{1}{3} \Rightarrow \alpha-\beta= \\ -\arctan \frac{1}{3} \text {. Therefore, } \beta=\arccos \frac{4}{5}+\arctan \frac{1}{3}, \cos \beta=\cos \left(\arccos \frac{4}{5}+\operatorname{arctan} \frac{2}{3}\right)=\frac{4}{5} \times \frac{3}{\sqrt{10}}- \\ \frac{3}{5} \times \frac{1}{\sqrt{10}}=\frac{9 \sqrt{10}}{50} .\end{array}$