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 ^{2} B=1-\frac{144}{169} \ & \therefore \sin ^{2} B=\frac{25}{169} \ &. \therefore \sin B=\frac{5}{13} \ & \therefore \cos (A+B)=\cos A \cos B-\sin A \sin B \end{aligned}

$ \begin{aligned} &=\left(\frac{4}{5}\right)\left(\frac{12}{13}\right)-\left(\frac{3}{5}\right)\left(\frac{5}{13}\right) \\ &=\frac{48}{65}-\frac{15}{65} \end{aligned} \\ \therefore \cos (A+B)=\frac{33}{65} \\ \therefore A+B=\cos ^{-1}\left(\frac{33}{65}\right) \\ \therefore \cos ^{-1}\left(\frac{4}{5}\right)+\cos ^{-1}\left(\frac{12}{13}\right)=\cos ^{-1}\left(\frac{33}{65}\right) $