Your argument for part a) is correct.
For b), think about where the problem occurs when we try to do similar steps. We can't prove uniform convergence immediately, because $x$ could be unbounded. So instead, for now just consider $x\in (-a,a).$ Then since $|\arctan x| \leq |x| $ we have $$\biggr|\sum_{n=1}^{\infty} \arctan \left( \frac{x}{n^2} \right) \biggr| \leq \sum_{n=1}^{\infty} \frac{a}{n^2} < \infty. $$
Thus the sum converges uniformly in $(-a,a).$
Now, the sequence of term by term derivatives converges uniformly in $(-a,a)$ as well since (as sos440 mentioned above) $$ \frac{d}{dx} \arctan \left( \frac{x}{n^2} \right) \leq \frac{1}{n^2}.$$
Thus, we conclude the sum is differentiable for all $x\in (a,-a).$ Since $a>0$ was arbitrary, the sum is differentiable for all $x\in \mathbb{R},$ with derivative $$ f'(x) = \sum_{n=1}^{\infty} \frac{1}{n^2 + (x/n)^2} .$$ In particular, $f'(0) = \pi^2/6 > 0.$
To clarify your confusion:
Weierstrass M-test states one and only one thing:
Given a sequence of functions $f_k(x)$ defined on $E\subseteq\mathbb{R}$, the series $\sum f_k(x)$ converges uniformly if there exists a sequnece of reals $M_k$ such that $|f_k|\le M_k$ for each $k$ and $\sum M_k$ converges.
Note here that $M_k$ must not depend on $x$. This only means that if you found such sequence $M_k$ then the series uniformly converges. It says nothing about the series if you have found some $M_k$ whose series does not converge.
In particular, this test cannot be used to prove that some series does not converge uniformly.
Now consider
$$
f_k(x)=\frac{1}{kx+2}-\frac{1}{(k+1)x+2}
$$
Consider the partial sum
\begin{align*}
S_n(x)=\sum_{k=0}^{n} f_k(x)&=\left(\frac{1}{2}-\frac{1}{x+2}\right)+\cdots+\left(\frac{1}{nx+2}-\frac{1}{(n+1)x+2}\right)\\
&=\frac{1}{2}-\frac{1}{(n+1)x+2}
\end{align*}
the series $\sum f_k(x)$ converges uniformly if and only if $S_n$ converges uniformly.
Now $S_n(0)=0$ for all $n$. So $S_n(0) \to 0$, and $S_n(x) \to \frac{1}{2}$ for $0<x\le 1$. So define
$$
S(x)=
\begin{cases}
0 & \text{if} \quad x=0 \\
\frac{1}{2} & \text{if} \quad 0<x\le 1
\end{cases}
$$
Then $S_n(x)\to S(x)$. Also, $S_n(0)-S(0)=0$ for any $n$. Now
$$
m_n:=\sup_{x\in [0,1]} |S_n-S|=\sup_{x\in (0,1]} \left\lvert -\frac{1}{(n+1)x+2} \right\rvert=\frac{1}{2} \not\to 0
$$
So $S_n$ does not converge uniformly.
Best Answer
Yes, since each finite sum is continuous, then the uniform convergence of the series implies the continuity of the limit on any compact of $\mathbb{R}$, thus the continuity of the limit sum on $\mathbb{R}$.
For the differentiability, you can check that the series $$ \left|\sum_{k=0}^{\infty}k\:e^{-k}\cos(kt)\right|\leq\left|\sum_{k=0}^{\infty}k\:e^{-k}\right|=\frac{e}{(1-e)^2}<+\infty $$ is (normally) uniformly convergent on $\mathbb{R}$, giving the desired the differentiability of the limit sum.