I have pretty much completed this question and have found the Fourier representation to be;
$$ f(x) =\frac A2 +\sum_{n=0}^\infty 2A\frac{\cos(((2n-1)(\pi x))/2f_o)}{\pi(2n-1)} $$
Now I don't understand what the question is asking for (b) and (c). Do I simply let Fourier series = 0 and then solve? For (b) then it would just be series converge to $-\frac{A}{2}$ = expand some of the series. And similarly for (c)?. Where would I go with that?
Let's see...the period of this even wave is $4 f_0$, so write
$$G(f) = G_0 + 2 \sum_{k=1}^{\infty} G_k \cos{\frac{2 \pi k f}{4 f_0}} $$
$$G_k = \frac{A}{4 f_0} \int_{-f_0}^{f_0} df \, \cos{\frac{2 \pi k f}{4 f_0}} = A \frac{\displaystyle \sin{\frac{\pi k}{2}}}{\pi k}$$
So
$$G(f) = \frac{A}{2} + 2 A \sum_{k=1}^{\infty} \frac{\displaystyle \sin{\frac{\pi k}{2}}}{\pi k} \cos{\frac{\pi k f}{2 f_0}} $$
A quick plot in Mathematica verified this form for $G$.
$$\begin{align}G(0) &= \frac{A}{2} + \frac{2 A}{\pi} \sum_{k=0}^{\infty} \frac{\displaystyle \sin{\frac{\pi k}{2}}}{k}\\ &= \frac{A}{2}+\frac{2 A}{\pi} \sum_{k=0}^{\infty}\frac{(-1)^k}{2 k+1}\\ &= \frac{A}{2}+\frac{2 A}{\pi} \frac{\pi}{4}\\ &= A \end{align}$$
For $f=f_0$ you may show that the series is instead
$$G(f_0) = \frac{A}{2} + \frac{A}{\pi} \sum_{k=1}^{\infty} \frac{\sin{\pi k}}{k} = \frac{A}{2}$$
You tell me: reasonable or unreasonable?