$\frac{\sin ^4 x}{2}+\frac{\cos ^4 x}{3}=\frac{1}{5}$
$\frac{3 \sin ^4 x+2 \cos ^4 x}{6}=\frac{1}{5}$
$\begin{aligned} & \Rightarrow 15 \sin ^4 x+10 \cos ^4 x=6 \\ & \Rightarrow 5 \sin ^4 x+10\left(\sin ^4 x+\cos ^4 x\right)=6 \\ & \Rightarrow 5 \sin ^4 x+10\left[\left(\sin ^2 x+\cos ^2 x\right)^2-2 \sin ^2 x \cos ^2 x\right]=6 \\ & \Rightarrow 5 \sin ^4 x+10\left(1-2 \sin ^2 x\left(1-\sin ^2 x\right)\right)=6 \\ & \Rightarrow 5 \sin ^4 x+10-20 \sin ^2 x+20 \sin ^4 x=6 \\ & \Rightarrow 25 \sin ^4 x-20 \sin ^2 x+4=0 \\ & \Rightarrow\left(5 \sin ^2 x-2\right)^2=0 \Rightarrow \sin ^2 x=\frac{2}{5}\end{aligned}$
$\begin{aligned} & \therefore \sec ^2 x=\frac{5}{3}, \operatorname{cosec}^2 x=\frac{5}{2} \\ & 27 \sec ^6 \alpha+8 \operatorname{cosec}^6 \alpha \\ & =27\left(\frac{5}{3}\right)^3+8\left(\frac{5}{2}\right)^3=27 \cdot \frac{125}{27}+8 \cdot \frac{125}{8}=250\end{aligned}$