I'm not sure if I'm doing better. Here is the stuff.
Consider the sequence of functions $$f_n(x) = nx \left(\frac{x}{n}\right)^n\text{sinc}^n\left(\frac{x}{n}\right)$$ and the series $$s(x) = \sum_{n=1}^{+\infty} f_n(x).$$
Point-wise convergence
Since $\text{sinc}(\alpha) \leq 1$, then $f_n(x) \leq nx \left(\frac{x}{n}\right)^n = g_n(x) $ and $$s(x) \leq \sum_{n=1}^{+\infty} g_n(x) = t(x).$$
$g_n(x)$ converges to $0$ as $n$ goes toward infinity. Using ratio test, I get that:
$$\lim_{n \to +\infty} \frac{g_{n+1}(x)}{g_n(x)} = \lim_{n \to +\infty}\frac{(n+1)^{-n}x^{n+2}}{n^{1-n}x^{1+n}} = \lim_{n \to +\infty} \frac{1}{n}\left(\frac{n+1}{n}\right)^{-n}x = 0 < 1.$$
Then I can state that the series $t(x)$ converges point-wise for all $x \in \mathbb{R}$. Since $s(x) \leq t(x)$, then also $s(x)$ converges point-wise for all $x \in \mathbb{R}$. Am I right?
Uniform convergence
I don't know where to start from...
Hint : $|f_n(x)|\le n |x| \left|\frac{x}{n} \right|^n =\frac{|x|^{n+1}}{n^{n-1}} \le \frac{|x|^{n+1}}{n!}$ and use the fonction $|x|e^{|x|}$
Edit :
Hint for uniform convergence : can you show that $$||s(x)-s_N(x)||_\infty \ge f_N(\frac{\pi N}{2})=\frac{N^2\pi}{2}$$ ? (You just have to prove that $\forall k >N, f_k(\frac{\pi N}{2}) \ge 0$ ) What does that mean ?