Solution: \begin{aligned} \cos ^{2} A &=1-\sin ^{2} A \ &=1-\left(\frac{5}{13}\right)^{2} \ &=1-\frac{25}{169}=\frac{144}{169}\&. \cos A=\pm \frac{12}{13}\ & \therefore \cos A=-\frac{12}{13} \quad(\because A \text { is obtuse angle }) \sin ^{2} B=1-\cos ^{2} B\ &=1-\left(-\frac{4}{5}\right)^{2}\& \sin ^{2} B=1-\frac{16}{25}=\frac{9}{25} \ & \sin B=\pm \frac{3}{5}\end{aligned}

\begin{aligned} \therefore \sin (A+B) &=\sin A \cdot \cos B+\cos A \cdot \sin B \ &=\left(\frac{5}{13}\right) \times\left(-\frac{4}{5}\right)+\left(-\frac{12}{13}\right) \times\left(\frac{3}{5}\right) \ &=-\frac{56}{65} \end{aligned}