Part (a):
\begin{equation*}
\hat{f}(\omega)=
\frac{1}{2\pi}\int_{-\infty}^\infty f(x)e^{-i\omega x}\,dx\\
= \frac{1}{2\pi}\int_{-a}^{a} f(x)e^{-i\omega x}\,dx\\
= \frac{1}{2\pi} \frac{e^{-\omega x}}{i\omega} \big|_{-a}^{a}\\
= \frac{i}{2\pi \omega}(e^{-i\omega a}-e^{i\omega a})\\
= \frac{e^{i\omega a}-e^{-i\omega a}}{2i\pi \omega}\\
=\frac{sin(\omega a)}{\pi \omega}.
\end{equation*}
Part (b):
Using the result from part (a) along with Theorem 3d:
\begin{equation*}
g = xf(x)\implies
\hat{g}(\omega) = i\hat{f}(\omega)\\
=i\frac{d}{d\omega}\big(\frac{sin(\omega a)}{\pi \omega} \big)\\
=i \frac{a\omega \cos{\omega a} - \sin{\omega a}}{\pi \omega^2}\\
=\frac{ia\cos{\omega a}}{\pi \omega} - \frac{i\sin{}\omega a}{\pi \omega^2}.
\end{equation*}
Part (c):
Let $$ f(x)=\left\{\begin{aligned} & 1&& |x|\le a,\\ & 0 && |x|> a;\end{aligned}\right.$$
Then using the result obtained from part (a), we have a fourier transform pair:
\begin{equation*}
f(x) = \int_{-\infty}^\infty \hat{f}(\omega)e^{i\omega x}\,d\omega\\
\implies f(x) = \int_{-\infty}^\infty \frac{\sin{\omega a}}{\pi \omega}e^{i\omega x}\,d\omega\\
\end{equation*}
Switch $\omega$ with $x$.
\begin{equation*}
\implies \left\{\begin{aligned} & \pi&& |\omega|\le a,\\ & 0 && |\omega|> a;\end{aligned}\right.= \int_{-\infty}^\infty \frac{\sin{x a}}{x}e^{i\omega x}\,dx\\
\end{equation*}
Now let $a = 1$, and $\omega = 0$. For these values the function gives us $\pi$.
Thus,
\begin{equation*}
\int_{-\infty}^\infty \frac{\sin{x}}{x}\,dx = \pi.
\end{equation*}
