Problem while solving this inequality. $$ \frac{1}{|x|-3} \lt \frac{1}{2}$$
$$|x|-3 \gt 2$$
$$|x| \gt 5$$
$$x\in(-\infty,-5)\cup(5,\infty)$$
But the answer given is: $$x\in(-\infty,-5)\cup(-3,3)\cup(5,\infty)$$
Where does the extra part of the solution set come from? What have I missed?
 A: A good way to see this is to notice that the function on the left is not continuous across its domain, but comes in three branches. When $x<-3$ or $x>3$, the left hand side is positive, and we can solve as you did. However, when $-3<x-3$, the left hand side is negative, so the inequality is satisfied for all $x$ in that part of the domain.
This is perhaps clearer with a simpler example: The inequality $\frac1x<2$ isn't just solved by $x>\frac12$, but also by all negative $x$.
To work analytically in a way that leads to noticing this, try moving everything to one side and making a common denominator:
$$\begin{align}
\frac{1}{|x|-3}-\frac12<0 &\implies \frac{2-|x|+3}{2(|x|-3)}<0\\
&\implies \frac{|x|-5}{|x|-3}>0
\end{align}$$
From this form, it is clear that we need to analyze the intervals $(-\infty,-5), (-5,-3),(-3,3),(3,5)$ and $(5,\infty)$ separately.
A: To solve inequalities, you need to find a set of solutions as well as show that all solutions are in that set.
That means for an inequality $f(x)<g(x)$ we need to find a set of solutions $S$ such that $f(x)<g(x)\iff x\in S$. So the best way to work with inequalities is to use equivalences rather than implications between lines. Alternatively, you can split the working into $\implies$ and $\impliedby$.
The problem in your working occurs as the first line doesn't even imply the second line. When working with inequalities, you want to apply operations to both sides so that the inequality is preserved.
Taking the reciprocal of both sides does not preserve this (unlike for equalities). If you consider the graph of $f(x)=\frac{1}{x}$, you'll see that for $a,b\in\Bbb R^*$, $a<b$ implies $f(a)<f(b)$ if $a$ and $b$ have different signs, and $f(a)>f(b)$ if $b$ and $a$ have same signs. Hence you need to split into different cases.
Alternatively, you can multiply both sides of the inequality by $2(|x|-3)$, but you still need to consider the different cases which arise from the sign of $|x|-3$.

You should instead have something like (note $\land$ and $\lor$ are logical 'and' and 'or' respectively):
$$\frac{1}{|x|-3} \lt \frac{1}{2}$$
$\iff$ (note $|x|-3\neq0$ so this is excluded on the next line)
$$(|x|-3>0\land|x|-3 \gt 2)\lor (|x|-3<0\land|x|-3 \lt 2)$$
$\iff$
$$(|x|>3\land|x| \gt 5)\lor (|x|<3\land|x| \lt 5)$$
$\iff$
$$|x| \gt 5\lor |x|<3$$
$\iff$
$$x\in(-\infty,-5)\cup(-3,3)\cup(5,\infty)$$
A: HINT: if we have $$x\geq 0$$ we get $$\frac{1}{x-3}<\frac{1}{2}$$ and now we get for $$x>3$$ the solution $$5<x$$ in the other case $$x<3$$ we get $$5>x$$
Can you finish?
A: Very simple: the rule
$$ A<B\iff \frac 1B>\frac1 A $$
is valid if and only if $A$ and $B$ have the same sign.
On the contrary,  if $A<0<B$, we have $\;\dfrac1Ax<0<\dfrac1B$, and precisely, 
$$\lvert x\rvert -3<0\iff -3<x<3.$$
