If the numerical value of tan $\left({\cos }^{-1} (4/5)\right.$$\left.+{\tan }^{-1} (2/3)\right)$ is a/b then

Let ${\cos }^{-1} (4/5)=\alpha$ that is, $\cos \alpha =4/5$, so that

$\tan \alpha =\sqrt{{\left(\frac{5}{4}\right)}^2-1}=\frac{3}{4}(\because 0<\alpha <\pi$ and $\cos \alpha >0)$

and $\tan \left({\cos }^{-1} \frac{4}{5}+{\tan }^{-1} \frac{2}{3}\right)=\frac{\tan \alpha +\frac{2}{3}}{1-\tan \alpha \cdot \frac{2}{3}}$

$=\frac{\frac{3}{4}+\frac{2}{3}}{1-\frac{2}{3}\cdot \frac{3}{4}}=\frac{17}{6}=\frac{a}{b}$   (given)

so $a=17,b=6,a+b=23,a-b=11$and  $3b=a+1$.