(a)
$$
\begin{aligned}
& I_n=\int_{-\pi}^\pi \frac{\sin n x}{\left(1+\pi^x\right) \sin x} d x \\
& I_n=\int_{-\pi}^\pi \frac{\pi^x \sin n x}{\left(1+\pi^x\right) \sin x} d x \\
& {\left[\because \int_a^b f(x) d x=\int_a^b f(a+b-x) d x\right]} \\
& 2 I_n=\int_{-\pi}^\pi \frac{\sin n x}{\sin x} d x
\end{aligned}
$$

$$
\begin{aligned}
& \Rightarrow \quad 2 I_n=2 \int_0^\pi \frac{\sin n x}{\sin x} d x \\
& \Rightarrow \quad I_n=\int_0^\pi \frac{\sin n x}{\sin x} d x \\
& \text { Now, } \quad I_{n+2}-I_n \\
& =\int_0^\pi \frac{\sin (n+2) x-\sin n x}{\sin x} d x \\
& =\int_0^\pi \frac{2 \cos (n+1) x \sin x}{\sin x} d x \\
& =2\left[\frac{\sin (n+1) x}{(n+1)}\right]_0^\pi=0 \\
& \Rightarrow \quad I_{n+2}=I_n \\
& \text { (b) As } I_3=I_5=\ldots=I_{21} \\
& \therefore \sum_{m=1}^{10} I_{2 m+1}=10 I_3=10 \int_0^\pi \frac{\sin 3 x}{\sin x} d x \\
& =10 \int_0^\pi\left(3-4 \sin ^2 x\right) d x \\
& =10[3 x-2 x+\sin 2 x]_0^\pi=10 \pi \\
& \text { (c) As } \quad I_2=I_4=\ldots=I_{20} \\
& \therefore \quad \sum_{m=1}^{10} I_{2 m}=10 \int_0^\pi \frac{\sin 2 x}{\sin x} d x \\
& =20[\sin x]_0^\pi=0 \\
&
\end{aligned}
$$