Given that,

$2{\tan }^{-1}\frac{1}{5}+{\sec }^{-1}\frac{5\sqrt{2}}{7}+2{\tan }^{-1}\frac{1}{8}$

$=2{\tan }^{-1}\frac{1}{5}+2{\tan }^{-1}\frac{1}{8}+{\sec }^{-1}\frac{5\sqrt{2}}{7}$

$=2\left({\tan }^{-1}\frac{1}{5}+{\tan }^{-1}\frac{1}{8}\right)+{\sec }^{-1}\frac{5\sqrt{2}}{7}$

$=2{\tan }^{-1}\left(\frac{{\displaystyle \frac{1}{5}}+{\displaystyle \frac{1}{8}}}{1-{\displaystyle \frac{1}{5}}·{\displaystyle \frac{1}{8}}}\right)+{\sec }^{-1}\frac{5\sqrt{2}}{7}$

$\left.\therefore {\tan }^{-1}A+{\tan }^{-1}B={\tan }^{-1}\left(\frac{A+B}{1-AB}\right)\right)$

$=2{\tan }^{-1}\left(\frac{8+5}{40-1}\right)+{\sec }^{-1}\frac{5\sqrt{2}}{7}$

$=2{\tan }^{-1}\left(\frac{13}{39}\right)+{\sec }^{-1}\frac{5\sqrt{2}}{7}$

$=2{\tan }^{-1}\left(\frac{1}{3}\right)+{\sec }^{-1}\frac{5\sqrt{2}}{7}$

$={\tan }^{-1}\left(\frac{2\times {\displaystyle \frac{1}{3}}}{1-{\left({\displaystyle \frac{1}{3}}\right)}^2}\right)+{\sec }^{-1}\frac{5\sqrt{2}}{7}$

$\left(\therefore 2{\tan }^{-1}A={\tan }^{-1}\left(\frac{2A}{1-A^2}\right)\right)$

$={\tan }^{-1}\left(\frac{6}{8}\right)+{\sec }^{-1}\frac{5\sqrt{2}}{7}$

$={\tan }^{-1}\left(\frac{3}{4}\right)+{\sec }^{-1}\frac{5\sqrt{2}}{7}$

$={\tan }^{-1}\left(\frac{3}{4}\right)+{\tan }^{-1}\left(\frac{1}{7}\right)$

$={\tan }^{-1}\left(\frac{{\displaystyle \frac{3}{4}}+{\displaystyle \frac{1}{7}}}{1-{\displaystyle \frac{3}{4}}·{\displaystyle \frac{1}{7}}}\right)$

$={\tan }^{-1}\left(\frac{21+4}{28-3}\right)$

$={\tan }^{-1}\left(\frac{25}{25}\right)$

$={\tan }^{-1}\left(1\right)$

$\therefore 2{\tan }^{-1}\frac{1}{5}+{\sec }^{-1}\frac{5\sqrt{2}}{7}+2{\tan }^{-1}\frac{1}{8}=\frac{\mathrm{\pi }}{4}$