4. Given $\alpha, \beta \in \mathbf{R}$, and $\cos \alpha+\cos \beta=\frac{1}{3}, \tan (\alpha+\beta)=\frac{24}{7}$, then $\sin \alpha+\sin \beta=$
$$
\cos \alpha+\cos \beta=2 \cos \frac{\alpha+\beta}{2} \cos \frac{\alpha-\beta}{2}=\frac{1}{3},
$$

Let $\sin \alpha+\sin \beta=2 \sin \frac{\alpha+\beta}{2} \cos \frac{\alpha-\beta}{2}=k$, dividing the two equations yields $\tan \frac{\alpha+\beta}{2}=3 k$.
Therefore, $\tan (\alpha+\beta)=\frac{2 \cdot 3 k}{1-9 k^{2}}=\frac{24}{7} \Rightarrow \frac{k}{1-9 k^{2}}=\frac{4}{7} \Rightarrow 36 k^{2}+7 k-4=0$ $\Rightarrow(9 k+4)(4 k-1)=0 \Rightarrow k=-\frac{4}{9}$ or $k=\frac{1}{4}$.