Solution: (\begin{aligned} \operatorname{Let} \cos ^{-1}\left(\frac{4}{5}\right) &=A \ \therefore \cos A &=\frac{4}{5} \ \therefore \sin ^{2} A &=1-\cos ^{2} A \ &=1-\cos ^{2} A \ &=\frac{9}{25} \ \therefore \sin A &=\frac{3}{5} \ & \cos ^{-1}\left(\frac{12}{13}\right)=B \ & \therefore \cos B=\frac{12}{13} \ & \therefore \sin ^{2} B=1-\cos ^{2} B\ & \therefore \sin …