First observe that each of the series converges pointwise on its given interval (using standard comparison tests and results on $p$-series, geometric series, and
alternating series.
Towards determining uniform convergence, let's first recall the Weierstrass $M$-test:
Suppose $(f_n)$ is a sequence of real-valued functions on the set $I$ and $(M_n)$ is a sequence
of positive real numbers such that $|f_n(x)|\le M_n$ for $x\in I$,
$n\in\Bbb N$. If the series $\sum M_n$ is convergent then $\sum f_n$
is uniformly convergent on $I$.
It is worthwhile to consider the heart of the proof of this theorem:
Under the given hypotheses, if $m>n$, then for any $x\in I$
$$\tag{1}
\bigl| f_{n+1}(x)+\cdots+f_m(x)\bigr|
\le| f_{n+1}(x)|+\cdots+|f_m(x)\bigr|
\le M_{n+1}+\cdots M_n.
$$
So if $\sum M_n$ converges, we can make the right hand side of $(1)$ as small as we wish. Noting that the right hand side of $(1)$ is independent of $x$, we can conclude that $\sum f_n$ is uniformly Cauchy on $I$, and thus uniformly convergent on $I$.
Now on to your problem:
To apply the $M$-test, you have to find appropriate $M_n$ for the series under consideration. Keep in mind that the $M_n$ have to be positive, summable, and bound the $|f_n|$. Sometimes they are easy to find, as in the series in a). Here note that
for any $n\ge 1$ and $x\in\Bbb R$,
$$
\biggl| {\sin(n^2x)\over n^2+x^2}\biggr|\le {1\over n^2}.
$$
So, take $M_n={1\over n^2}$ and apply the $M$-test. The series in a) converges uniformly on $\Bbb R$.
Sometimes finding the $M_n$ is not so easy. This is the case in c). Crude approximations for $f_n(x)=x^2e^{-nx}$ will not help. However, we could try to find the maximum value of $f_n$ over $(0,\infty)$ and perhaps this will give us what we want. And indeed, doing this (using methods from differential calculus), we discover that
the maximum value of $f_n(x)=x^2e^{-nx}$ over $(0,\infty)$ is ${4e^{-2}\over n^2}$. And now the road towards using the $M$-test is paved...
Sometimes the $M$-test doesn't apply. This is the case for the series in b), the required $M_n$ can't be found (at least, I can't find them).
However, here, the proof of the $M$-test gives us an idea. Since the series in b) is alternating (that is, for each $x\in[0,\infty)$, the series $\sum\limits_{n=1}^\infty{(-1)^n\over x+n}$ is alternating), perhaps we can show it is uniformly Cauchy on $[0,\infty)$.
Indeed we can:
For any $m\ge n$ and $x\ge0$
$$\tag{2}
\Biggl|\,{(-1)^n\over n+x}+{(-1)^{n+1}\over (n+1)+x}+\cdots+{ (-1)^m\over m+x}\,\Biggl|\ \le\ {1\over n+x}\le {1\over n}.
$$
The term on the right hand side of $(2)$ is independent of $x$ and can be made as small as desired. So, the series in b) is uniformly Cauchy on $[0,\infty)$, and thus uniformly convergent on $[0,\infty)$.
By well known Cauchy–Hadamard theorem for power series $\sum\limits_{n=0}^{\infty}(z-z_0)^nc_n$ we have, that so called convergence radius $\frac{1}{R}=\lim\limits_{n \to \infty}\sup\sqrt[n]{|c_n|}$. In our case
$$\sqrt[n]{\frac{e^{-n}}{\sqrt{n}}} \to \frac{1}{e}=\frac{1}{R}$$
So we have pointwise convergence for $|x|<e$. In right border point we have divergence as for $\frac{1}{\sqrt{n}}$ and for left convergence $\frac{(-1)^n}{\sqrt{n}}$.
As it is known, uniform convergence we have on each closed segment within convergence interval.
Best Answer
Hints
For your first question: Weirestrass $\;M$-test;
For your second question: no. For example, the series of $\,\frac1{1-x}\;$ converges only for $\,|x|<1\,$ , and the convergence is uniform for $\,|x|\leq<1\;$ ...