Given $ h(x)=f(x)+O(g(x)) $ and knowing that $ \lim_{x \to \infty}=\frac{g(x)}{f(x)}=0$ (int other words $f(x)=o(g(x))$) find such F(x) and G(x), $\frac{1}{h(x)}=F(x)+O(G(x)) $.
Because $ f(x)=o(g(x)) $ that for any $e>0$ we can find $n_{0}$ such for every $n>n_{0}$ $g(n)<e\cdot f(n)$ . And knowing that $ h(x)=f(x)+O(g(x)) \leq f(x)+ c \cdot g(x) $ for some $ c>0 $ for all but finite amount of $x$'s we take $ e=\frac{1}{c} $ from the limit and have that $h(x)=f(x)+O(g(x)) \leq f(x)+f(x) \Rightarrow \frac{1}{h(x)}>= \frac{1}{2 \cdot f(x)}$.
But this has gotten me completely nowhere and I don't know that to do...
You have $\frac{h(x)}{f(x)}=1+O(\frac{g(x)}{f(x)})$. Now note that $\lim \frac{g(x)}{f(x)}=0$ implies that the reciprocal is also $\frac{f(x)}{h(x)}=1+O(\frac{g(x)}{f(x)})$, hence $$\frac{1}{h(x)} = \frac1{f(x)}+O\left(\frac{g(x)}{f^2(x)}\right).$$
In case you didn't know the helpful result I used:
Lemma. If $F(x)=1+O(G(x))$ with $G(x)\to 0$, then $\frac1{F(x)} = 1+O(G(x))$.
Proof: As $F(x)=1+O(G(x))$, there is some $x_0$ such that $|F(x)-1|<cG(x)$ for $x>x_0$. And by the second condition, there exists $x_1$ such that $|G(x)|<\frac1{2c}$ for $x>x_1$. Hence $|F(x)-1|<\frac12$ and then $F(x)>\frac12$ for $x>x_2:=\max\{x_0,x_1\}$. Therefore $$\left|\frac1{F(x)}-1\right|=\frac{|1-F(x)|}{|F(x)|}<2cG(x)$$ for all $x>x_2$. $_\square$
Note that we are not "wasting" anything in the conclusion because when starting with $\frac1{F(x)}=1+O(G(x))$ and applying the lemma again, we obtain our original $F(x)=1+O(G(x))$ back.