SOLUTION : $\text { Here } f(x)=\left(3-x^3\right)^{1 / 3} \Rightarrow(\text { fof }) x=f\left(f(x)=f\left(\left(3-x^3\right)^{1 / 3}\right)\right. \\$
$=\left[3-\left(\left(3-x^3\right)^{1 / 3}\right)^3\right]^{1 / 3}=\left(3-3+x^3\right)^{1 / 3}=\left(x^3\right)^{1 / 3}=x \\$
$\text { Let } x=\cos ^{-1} \frac{4}{5} \Rightarrow \cos x=\frac{4}{5} \\$
$\therefore \sin x=\sqrt{1-\cos ^2 x}=\sqrt{1-\frac{16}{25}}=\sqrt{\frac{9}{25}}=\frac{3}{5} \text { and } y=\cos ^{-1} \frac{12}{13} \Rightarrow \cos y=\frac{12}{13} \\$
$\therefore \sin y=\sqrt{1-\cos ^2 y}=\sqrt{1-\frac{144}{169}}=\sqrt{\frac{25}{169}}=\frac{5}{13} \\$
$\Rightarrow \cos (x+y)=\cos ^x \cos y-\sin x \sin y=\frac{4}{5} \cdot \frac{12}{13}-\frac{3}{5} \cdot \frac{5}{13}=\frac{48}{65}-\frac{15}{65}=\frac{33}{65} \\$
$\therefore \quad x+y=\cos ^{-1}\left(\frac{33}{65}\right) \text {. Hence } \cos ^{-1}\left(\frac{4}{5}\right)+\cos ^{-1}\left(\frac{12}{13}\right)=\cos ^{-1}\left(\frac{33}{65}\right)$
