If $\large\frac{\sin^4\theta}{2}+\frac{\cos^4\theta}{2}=\frac{1}{5}$ then

$\begin{array}{1 1}(a)\;\tan^2x=\large\frac{2}{3}&(b)\;\large\frac{\sin^8 x}{8}+\frac{\cos^8x}{27}=\frac{1}{125}\\(c)\;\tan^2x=\large\frac{1}{3}&(d)\;\large\frac{\sin^8x}{8}+\frac{\cos^8 x}{27}=\frac{2}{125}\end{array}$

Given :
$\large\frac{\sin^4x}{2}+\frac{\cos^4x}{3}=\frac{1}{5}$
$\Rightarrow \large\frac{\sin^4x}{2}+\frac{(1-\sin^2x)^2}{3}=\frac{1}{5}$
$\Rightarrow \large\frac{\sin^4x}{2}+\frac{1+\sin^4x-2\sin^2x}{3}=\frac{1}{5}$
$\Rightarrow \large\frac{3\sin^4x+2+2\sin^4x-4\sin^2x}{6}=\frac{1}{5}$
$\Rightarrow 5\sin^4x-4\sin^2x+2=\large\frac{6}{5}$
$\Rightarrow 25\sin^4x-20\sin^2x+4=0$
$\Rightarrow (5\sin^2x-2)^2=0$
$\Rightarrow \sin^2x=\large\frac{2}{5}$
$\cos^2x=\large\frac{3}{5}$
$\tan^2x=\large\frac{2}{3}$
$\Rightarrow \large\frac{\sin^8x}{8}+\frac{\cos^8x}{27}=\frac{1}{125}$
Hence (b) is the correct answer.