I'll tackle the Frobenius elements first.
Let $K=\mathbb{Q}(i,\sqrt{5})$. We have $\operatorname{Gal}(K/\mathbb{Q})=\{\operatorname{id}_K,\rho,\tau,\rho\tau\}\cong\mathbb{Z}/2\mathbb{Z}\times\mathbb{Z}/2\mathbb{Z}$, where $\rho$ is complex conjugation and $\tau$ is the automorphism of $K$ switching $\sqrt{5}$ and $-\sqrt{5}$.
Fact 1: Given
- an extension of number fields $E/F$
- a prime $P\subset\mathcal{O}_F$ that doesn't ramify in $E$
- any prime $\mathcal{P}\subset\mathcal{O}_E$ above $P$
then for any intermediate field $L$ such that $L/F$ is Galois, letting $\mathbf{P}=\mathcal{P}\cap\mathcal{O}_L$ the prime of $L$ lying beneath $\mathcal{P}$, we have $(\mathcal{P},E/F)|_L=(\mathbf{P},L/F).$
Fact 2: For a squarefree integer $D$, the field $M=\mathbb{Q}(\sqrt{D})$ has discriminant $d_M=4D$ if $D\equiv 2,3\bmod 4$ and $d_M=D$ if $D\equiv 1\bmod 4$, and if $p\nmid d_M$ is an odd prime, $p\text{ splits if }\left(\frac{D}{p}\right)=1\qquad\text{and}\qquad p\text{ is inert if }\left(\frac{D}{p}\right)=-1,$ (these being Legendre symbols).
I can include proofs of these facts if you would like.
Let $p$ be a prime of $\mathbb{Q}$ that doesn't ramify in $K$, i.e. $p\neq 2,5$. By Fact 1, $\textstyle\left.\left(\frac{K/\mathbb{Q}}{p}\right)\right|_{\mathbb{Q}(\sqrt{5})}=\left(\frac{\mathbb{Q}(\sqrt{5})/\mathbb{Q}}{p}\right)\in\operatorname{Gal}(\mathbb{Q}(\sqrt{5})/\mathbb{Q})\quad\text{and}\quad\left.\left(\frac{K/\mathbb{Q}}{p}\right)\right|_{\mathbb{Q}(i)}=\left(\frac{\mathbb{Q}(i)/\mathbb{Q}}{p}\right)\in\operatorname{Gal}(\mathbb{Q}(i)/\mathbb{Q})$ Thus, computing $\left(\frac{\mathbb{Q}(\sqrt{5})/\mathbb{Q}}{p}\right)$ and $\left(\frac{\mathbb{Q}(i)/\mathbb{Q}}{p}\right)$ will allow us to find $\left(\frac{K/\mathbb{Q}}{p}\right)$, because an element of $\operatorname{Gal}(K/\mathbb{Q})$ is determined by what it does to $\sqrt{5}$ and $i$. Let $\mathfrak{P}\subset\cal{O}_{\mathbb{Q}(\sqrt{5})}$ and $\mathfrak{p}\subset\cal{O}_{\mathbb{Q}(i)}$ be any primes lying over $p$.
There are two elements of $\operatorname{Gal}(\mathbb{Q}(\sqrt{5})/\mathbb{Q})$, $\operatorname{id}_{\mathbb{Q}(\sqrt{5})}$ and $\tau$, where $\tau(\sqrt{5})=-\sqrt{5}$. Note that $\left(\frac{\mathbb{Q}(\sqrt{5})/\mathbb{Q}}{p}\right)=\operatorname{id}_{\mathbb{Q}(\sqrt{5})}$ if and only if $\operatorname{id}_{\mathbb{Q}(\sqrt{5})}$ induces the Frobenius automorphism on $(\cal{O}_{\mathbb{Q}(\sqrt{5})}/\mathfrak{P})/(\mathbb{Z}/p\mathbb{Z})$, i.e. every element of $\cal{O}_{\mathbb{Q}(\sqrt{5})}/\mathfrak{P}$ is its own $p$th power, which is the case if and only if $\cal{O}_{\mathbb{Q}(\sqrt{5})}/\mathfrak{P}\cong\mathbb{F}_p$, i.e. $f(\mathfrak{P}/p)=1$. Recall that $p$ does not ramify in $\mathbb{Q}(\sqrt{5})$ by assumption, so that $e(\mathfrak{P}/p)=1$. Because $\mathbb{Q}(\sqrt{5})/\mathbb{Q}$ is Galois of prime degree, $e(\mathfrak{P}/p)=1$ and $f(\mathfrak{P}/p)=1$ if and only if $p$ splits in $\mathbb{Q}(\sqrt{5})$. Thus, we have shown that $\left(\frac{\mathbb{Q}(\sqrt{5})/\mathbb{Q}}{p}\right)=\operatorname{id}_{\mathbb{Q}(\sqrt{5})}$ if and only if $p$ splits in $\mathbb{Q}(\sqrt{5})$, which Fact 2 tells us happens if and only if $\left(\frac{5}{p}\right)=1$. By quadratic reciprocity, this is the case if and only if $\left(\frac{p}{5}\right)=1$, i.e. $p\equiv1,4\bmod 5$. Thus $\left(\frac{\mathbb{Q}(\sqrt{5})/\mathbb{Q}}{p}\right)=\begin{cases}\operatorname{id}_{\mathbb{Q}(\sqrt{5})}\quad\text{ if }p\equiv1,4\bmod 5\\ \,\,\,\tau\quad\quad\quad\quad\;\text{ if }p\equiv 2,3\bmod 5\end{cases}$
A similar analysis with $\frak{p}$ shows that $\left(\frac{\mathbb{Q}(i)/\mathbb{Q}}{p}\right)=\begin{cases}\operatorname{id}_{\mathbb{Q}(i)}\quad\text{ if }p\equiv1\bmod 4\\ \,\,\,\rho\quad\quad\quad\;\text{ if }p\equiv 3\bmod 4\end{cases}$
Because
$\begin{array}{c|c|c} & p\equiv1\bmod4 & p\equiv 3\bmod 4 \\\hline p\equiv 1,4\bmod 5 & p\equiv 1,9\bmod 20 & p\equiv11,19\bmod 20\\\hline p\equiv 2,3\bmod 5 & p\equiv 17,13\bmod 20 & p\equiv 7,3\bmod 20\\\hline \end{array}$ we have that $\left(\frac{K/\mathbb{Q}}{p}\right)=\begin{cases}\operatorname{id}_K\quad\text{ if } p\equiv 1,9\bmod 20\\ \tau\quad\quad\;\;\text{ if } p\equiv 13,17\bmod20 \\ \rho\quad\quad\;\;\text{ if } p\equiv 11,19\bmod 20\\ \rho\tau\quad\quad\text{ if }p \equiv 3,7\bmod 20\end{cases}$
Now to find the decomposition and inertia groups for $2$ and $5$. We will need
Fact 3: Given
- an abelian Galois extension of number fields $L/F$
- a prime $P\subset\mathcal{O}_F$
- a prime $\mathcal{P}\subset\mathcal{O}_L$ lying over $P$
then
- the inertia field of $\mathcal{P}$ over $P$, i.e. the fixed field of $I(\mathcal{P}/P)$, is the largest intermediate field in which $P$ does not ramify
- the decomposition field of $\mathcal{P}$ over $P$, i.e. the fixed field of $D(\mathcal{P}/P)$, is the largest intermediate field in which $P$ splits completely.
The intermediate fields of $K/\mathbb{Q}$ are, of course, $K$ $\text{ / }\qquad |\qquad \text{ \ }$ $\quad\quad\mathbb{Q}(\sqrt{5})\quad\mathbb{Q}(i\sqrt{5})\,\,\,\,\quad \mathbb{Q}(i)\quad\quad$ $\text{ \ } \qquad |\qquad \text{ / }$ $\mathbb{Q}$ which have discriminants $\begin{array}{ccccc} & & 400 & & \\ & & & & \\ 5 & & -20 & & -4\\ & & & & \\ & & 1 & & \end{array}$ (sorry about the ugly diagram, the LaTeX on the site has some limitations)
Because a prime ramifies if and only if it divides the discriminant, we see that the inertia field of $2$ is $\mathbb{Q}(\sqrt{5})$, and the inertia field of $5$ is $\mathbb{Q}(i)$. Thus, $I(\mathcal{P}/2)=\{\operatorname{id}_K,\rho\}$ for any prime $\mathcal{P}\subset\mathcal{O}_K$ lying over $2$, and $I(\mathfrak{P}/5)=\{\operatorname{id}_K,\tau\}$ for any prime $\mathfrak{P}\subset\mathcal{O}_K$ lying over $5$.
Because the decomposition group contains the inertia group, the inertia field contains the decomposition field. Thus, the decomposition field for $2$ is either $\mathbb{Q}(\sqrt{5})$ or $\mathbb{Q}$, and the decomposition field for $5$ is either $\mathbb{Q}(i)$ or $\mathbb{Q}$. By Fact 2, $2$ is inert in $\mathbb{Q}(\sqrt{5})$ and $5$ splits in $\mathbb{Q}(i)$ (indeed $5=(2+i)(2-i)$), so the decomposition field for $2$ is $\mathbb{Q}$ and the decomposition field for $5$ is $\mathbb{Q}(i)$. Thus, $D(\mathcal{P}/2)=\{\operatorname{id}_K,\rho,\tau,\rho\tau\}=\operatorname{Gal}(K/\mathbb{Q})$ for any prime $\mathcal{P}\subset\mathcal{O}_K$ lying over $2$, and $D(\mathfrak{P}/5)=\{\operatorname{id}_K,\tau\}$ for any prime $\mathfrak{P}\subset\mathcal{O}_K$ lying over $5$.