Step 1:
$A=2\tan^{-1}(2\sqrt 2-1)$
$\sqrt 2=1.414$
$\;\;=2\tan^{-1}(2\times 1.414-1)$
$\;\;=2\tan^{-1}(1.828)$
$\tan 60^{\large\circ}=\sqrt 3$
$\Rightarrow 1.732$
$\tan^{-1}(1.732)=60^{\large\circ}$
$\;\;=2\times ( >60^{\large\circ})$
$\;\;=>120^{\large\circ}$
Step 2:
$B=3\sin^{-1}\big(\large\frac{1}{3}\big)$$+\sin^{-1}\big(\large\frac{3}{5}\big)$
$\;\;=\sin^{-1}\big[3\times \large\frac{1}{3}$$-4\big(\large\frac{1}{3}\big)^3\big]+$$\sin^{-1}\big(\large\frac{3}{5}\big)$
$\;\;=\sin^{-1}\big(\large\frac{23}{27}\big)$$+\sin^{-1}\big(\large\frac{3}{5}\big)$
$\;\;=\sin^{-1}(0.852)+\sin^{-1}(0.60)$
$\;\;= (< 60^{\large\circ})+( < 45^{\large\circ})$
$\;\;=< 105^{\large\circ}$
$\Rightarrow A > B$
Hence (c) is the correct answer.