(a)

Given, $\ln f(x)=10 \ln \sin x+$ $\ln \left(1+\cos ^2 x+\cos ^4 x+\cos ^8 x\right)$

$\frac{f^{\prime}(x)}{f(x)}=10 \cot x-\sin 2 x$

${\left[\frac{1+2 \cos ^2 x+4 \cos ^6 x}{1+\cos ^2 x+\cos ^4 x+\cos ^8 x}\right] }$

$\left[\frac{1+2 \cos ^2 x+4 \cos ^6 x}{1+\cos ^2 x+\cos ^4 x+\cos ^8 x}\right]$

$\because f(x)$ is periodic with period $\pi$ and $\frac{f^{\prime}(x)}{f(x)}$ is continuous in $(0, \pi) .$

$\frac{f^{\prime}(x)}{f(x)}=\underbrace{10 \cot x}_{\begin{array}{c}\text { Range is }(-\infty, \infty) \\ \text { for } x \in(0, \pi)\end{array}}$

$\underbrace{-\sin 2 x\left[\frac{1+2 \cos ^2 x+4 \cos ^6 x}{1+\cos ^2 x+\cos ^4 x+\cos ^8 x}\right]}_{\text {finite number for } x \in(0, \pi)}$

So, $\frac{f^{\prime}(x)}{f(x)}$ has range $(-\infty, \infty)$ Hence, $\lambda \in R$.