(a) You are given $f(x)$ for $-\pi\leq x\leq \pi$. Hence you can sketch the function over $[-\pi,\pi]$. The condition $f(x+2\pi)=f(x)$ means that $f$ is $2\pi$-periodic. So to sketch $f$ over three periods means to sketch $f$ on an interval of length $3\cdot2\pi=6\pi$ and you can do so by replicating the sketch on $[-\pi,\pi]$. What you get is:
(b) To obtain the Fourier series of $f$ we must calculate its Fourier coefficients. In complex exponential form, the $k$-th coefficient is
$$
\hat{f}(k) = \frac{1}{2\pi} \int_{-\pi}^{\pi}f(t)e^{-ikt} \,dt
$$
Here because $f$ is given piecewise we split the integral in two:
\begin{align}
\hat{f}(k) &= \frac{1}{2\pi} \int_{-\pi}^{\pi}f(t)e^{-ikt} \,dt \\
&= \frac{1}{2\pi}\left( \int_{-\pi}^{0} (t-\pi)e^{-ikt} \,dt+\int_{0}^{\pi} (\pi-t)e^{-ikt} \,dt \right)
\end{align}
It is a simple calculus exercise to evaluate these integrals (do it by integration by parts for the $te^{-ikt}$ components). Don't forget to treat the cases $k=0$ and $k\neq0$ separately. The result is:
\begin{align}
\hat{f}(k) &= \begin{cases}\frac{\pi}{2}&&k=0\\ \frac{1+e^{-i\pi k}}{\pi k^2} && k\neq0\end{cases} \\
&=\begin{cases}\frac{\pi}{2}&&k=0\\ \frac{1+(-1)^{k+1}}{\pi k^2} && k\neq0\end{cases}
\end{align}
The Fourier series of $f$ is then
$$
\sum_{k=-\infty}^{\infty}\hat{f}(k)e^{ikt}
$$
(c) Since $f$ is continuous at $0$ and $C^1$ in a punctured neighborhood of $0$, we know the symmetric partial sums of the Fourier series of $f$ converge to $f(0)$. That is,
$$
\lim_{n\to\infty}\sum_{k=n}^{n}\hat{f}(k) = f(0)
$$
which, in this case, becomes
$$
\frac{\pi}{2} + \lim_{n\to\infty}\sum_{\substack{-n\leq k \leq n \\ k\neq0}}\frac{1+(-1)^{k+1}}{\pi k^2} = \pi\tag{1}
$$
Since $\displaystyle\frac{1+(-1)^{k+1}}{\pi k^2}$ is even, $(1)$ reduces to
$$
\frac{\pi}{2} + 2 \lim_{n\to\infty}\sum_{k=1}^{n}\frac{1+(-1)^{k+1}}{\pi k^2} = \pi\tag{2}
$$
Finally, we note that the numerator $1+(-1)^{k+1}$ equals $0$ when $k$ is even and $2$ when $k$ is odd. So $(2)$ reduces to
$$
\frac{\pi}{2} + \frac{4}{\pi} \sum_{k=1}^{\infty}\frac{1}{(2k-1)^2} = \pi\tag{3}
$$
The series in $(3)$ is precisely the one we are asked to evaluate. From $(3)$ we obtain
$$
\sum_{k=1}^{\infty}\frac{1}{(2k-1)^2} = \frac{\pi^2}{8}
$$
as was to be shown.
