Specify the convergence of function series The task is to specify convergence of infinite function series (pointwise, almost uniform and uniform convergence):
a) $\displaystyle \sum_{n=1}^{+\infty}\frac{\sin(n^2x)}{n^2+x^2}, \ x\in\mathbb{R}$
b) $\displaystyle\sum_{n=1}^{+\infty}\frac{(-1)^n}{n+x}, \ x\in[0;+\infty)$
c) $\displaystyle\sum_{n=0}^{+\infty}x^2e^{-nx}, \ x\in(0;+\infty)$
I know basic facts about these types of convergence and Weierstrass M-test, but still have problems with using this in practise.. 
 A: 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 $[0,\infty)$.
