The idea is to use the fact that $|a_n-L|$ can be made arbitrarily small to show $||a_n|-|L||$ can be made arbitrarily small. In this case, you can use that for each $x,y$, $$||x|-|y||\leq |x-y|$$
The second case is not always true: you need $L\neq 0$. Now note
$$\left| {\frac{1}{{{a_n}}} - \frac{1}{L}} \right| = \left| {\frac{{L - {a_n}}}{{{a_n}L}}} \right| = \frac{{\left| {{a_n} - L} \right|}}{{\left| {{a_n}L} \right|}}$$
Since $a_n\to L$, there is an $n_0$ such that $|a_n-L|<|L|/2$ whenever $n\geq n_0$. Then
$$|L|-|a_n|\leq |a_n-L|<|L|/2$$
so that $|a_n|>|L|/2$. Again, since $a_n\to L$, there is a $n_1$ for which
$$|a_n-L|<\epsilon |L|^2/2$$ whenever $n\geq n_1$. Then, for $n\geq \max\{n_0,n_1\}$, we must have
$$\left| {\frac{1}{{{a_n}}} - \frac{1}{L}} \right| = \frac{{\left| {{a_n} - L} \right|}}{{\left| {{a_n}L} \right|}} < \frac{2}{{\left| L \right|}}\frac{{\left| {{a_n} - L} \right|}}{{\left| L \right|}} < \frac{2}{{\left| L \right|}}\frac{{\varepsilon {{\left| L \right|}^2}}}{{2\left| L \right|}} = \varepsilon $$
Note there is some unusual steps involved, but you get the idea.