7. Given $\cos \alpha+\cos \beta=\frac{\sqrt{2}}{4}, \tan (\alpha+\beta)=\frac{4}{3},$ the value of $\sin \alpha+\sin \beta$ is . $\qquad$
7. $-\frac{\sqrt{2}}{2}$ or $\frac{\sqrt{2}}{8}$.

Let $\sin \alpha + \sin \beta = m$, then $\frac{\sin \alpha + \sin \beta}{\cos \alpha + \cos \beta} = \tan \frac{(\alpha + \beta)}{2} = \frac{m}{\frac{\sqrt{2}}{4}} = 2 \sqrt{2} m$. And $\tan (\alpha + \beta) = \frac{2 \tan \frac{(\alpha + \beta)}{2}}{1 - \tan^2 \frac{(\alpha + \beta)}{2}}$, so $\frac{4}{3} = \frac{4 \sqrt{2} m}{1 - 8 m^2}$. Therefore, $8 m^2 + 3 \sqrt{2} m - 1 = 0, m = -\frac{\sqrt{2}}{2}$ or $m = \frac{\sqrt{2}}{8}$.