For $(a)$ I got $$u(x,t) = \sum_{n=1}^{\infty} \frac{8}{\pi^3 n^3} e^{(\alpha - \pi^2 n^2)t} \sin(\pi n x)$$
Which I'm certain is correct.
Now for $(b)$, I know that if $x$ is an integer, then the result is $0$. If $x$ is not an integer: if $\alpha <\pi^2 n^2$ then the result is $0$ and if $\alpha >=\pi^2 n^2$ then the result is infinity. But the part I'm having issues is showing that $u(x,t)$ is uniform convergent, I have no idea how to do this and would appreciate some help.
Observe \begin{align} \left|\sum^\infty_{n=1} \frac{8}{\pi^3n^3}e^{(\alpha-\pi^2n^2)t}\sin(\pi n x)\right| \leq \frac{8}{\pi^3}\sum^\infty_{n=1}\frac{1}{n^3} \end{align} provided $t \in (0, \infty)$ and $\alpha-\pi^2\leq 0$, then by Wierestrass's M-test we have that the original series converges uniformly for all $t \in(0, \infty)$.
In the other case when $\alpha-\pi^2\geq 0$, then we see that \begin{align} \left|\sum^\infty_{n=1} \frac{8}{\pi^3n^3}e^{(\alpha-\pi^2n^2)t}\sin(\pi n x)\right| \leq \frac{8e^{\alpha t}}{\pi^3}\sum^\infty_{n=1}\frac{1}{n^3} \leq \frac{8e^{\alpha T}}{\pi^3}\sum^\infty_{n=1}\frac{1}{n^3} \end{align} for all fixed $T>0$. Hence the series converges uniformly on any compact interval $[0, T]$.