A starter would be the observation that prime ideals of an extension field are always above prime ideals of the ground field. In this case, the extension field is $F$, and the ground field is $\mathbb Q$, where all prime ideals are principal ideals generated by prime numbers. Therefore we are reduced to looking for the prime ideal factorisations of primes in the extension $F$.
By the theorem of Dedekind, we can find the prime ideal factorisation of $p\in \mathbb Z$ by examining the factorisation of the polynomial $f=x^2+5$ modulo $p$. If $p=5$, then $(p)$ ramifies in $F$, and is equal to $\mathfrak p^2$ for some prime ideal $\mathfrak p$ of $F$. In fact, as $5=-\sqrt{-5}^2$, we know that $(5)=(\sqrt{-5})^2$. And if $p=2$, then $x^2+5\equiv(x+1)^2\pmod2$ shows that $(2)$ ramifies as well: $(2)=(1+\sqrt{-5})^2$.
Now, if $p\ne2,5$, then $(p)$ is unramified in $F$, so either $p$ is inert, i.e. $(p)$ is still a prime ideal in $F$, or $p$ splits, meaning that $(p)=\mathfrak p_1\mathfrak p_2$. And a prime $p$ splits if and only if $x^2+5\equiv 0\pmod p$ has solutions in integers. Namely, if and only $(\dfrac{-5}{p})=1$. By quadratic reciprocity, we find that this is thus equivalent with $p\equiv 1, 3, 7, 9\pmod{20}$.
In summary, $\begin{cases}(p)=(\sqrt{-5})^2 \text{ if }p=5 \\ (p)=\text{ still prime if }p\equiv 11, 13, 17, 19\pmod{20}\\ (p)=\mathfrak p_1\mathfrak p_2 \text{ if } p\equiv1, 3, 7, 9\pmod{20}\end{cases}.$
So all prime ideals of $F$ are classified now.
As to the second question, I cannot understand properly the content of the question, as we know that the valuations correspond to the prime ideals, and hence information about one is information about the other. As a further detail, consider the localisations at various primes: this should make clear what the corresponding valuations are.
Please, inform me of errors.
Thanks for the attention.