(c) By Parseval's Theorem we have $\sum_{n=1}^{\infty} \frac{sin^2(n\delta)}{n^2\delta} = \frac{\pi-\delta}{2} $. (d) Let $\delta \rightarrow 0$ and prove that $\int_0^\infty (\frac{sinx}{x})^2dx = \frac{\pi}{2} $
I've shown that the improper integral exists and that the $N^{th}$ partial sum of the left hand side of (c) taking $\delta = \delta_N = \frac{R(\epsilon)}{N}$ is a Riemann sum where $R(\epsilon)$ is a number large enough so that the integral from 0 to $R(\epsilon)$ is close enough to the improper integral. Thus $\sum_{n=1}^{N}\frac{sin^2(n\delta_N)}{n^2\delta_N} $ is close enough to the improper integral for N large enough.
But I don't see how we can choose N large enough s.t $\sum_{n=1}^{N}\frac{sin^2(n\delta_N)}{n^2\delta_N} $ should be close to $\frac{\pi-\delta_N}{2}$ because I think the convergence of the infinite series is dependent on $\delta$.
Try using uniform convergence. Taking $N\in\mathbb{N}$ and $\delta$ of the form $1/k, k\in\mathbb{N}$ for convenience, $$\int_0^{\infty} \left(\frac{\sin x}{x}\right)^2 dx=\lim_{N\rightarrow \infty} \int_0^N \left(\frac{\sin x}{x}\right)^2 dx = \lim_{N\rightarrow \infty} \lim_{\delta\rightarrow 0} \sum_{n=1}^{N/\delta} \frac{\sin^2n\delta}{n^2\delta},$$ and we have the desired result if we could change the orders of the two limits. Recall from Chapter 7 that this is possible if $g_N(\delta):= \sum_{n=1}^{N/\delta} \frac{\sin^2n\delta}{n^2\delta}$ converges uniformly to $g(\delta):=\frac{\pi-\delta}{2}$, on some set having 0 as a limit point, in this case $\{1, \frac{1}{2}, \frac{1}{3},\dots\}$. The proposition follows from $$ |g(\delta)-g_N(\delta)|=\sum_{n=N/\delta+1}^{\infty} \frac{\sin^2n\delta}{n^2\delta} \leq \sum_{n=N/\delta+1}^{\infty} \frac{1}{n(n-1)\delta}=\frac{1}{N}.$$