Let me briefly sketch what I did (hope there is no mistake):
Let $L$ be the galois extension of $x^5+2$ over $\mathbb{Q}$.
Let $a$ be symbol such that $a^5 = -2$ and $\xi$ be $5^{th}$ primitive root of unity.
Now, $\{a, \xi a, \xi^2 a, \xi^3 a, \xi^4 a \}$ are roots of $x^5+2$ and $L = \mathbb{Q}(a,\xi)$.
Since $x^5+2$ is irreducible, $[\mathbb{Q}(a),\mathbb{Q}]=5$
and $[\mathbb{Q}(\xi):\mathbb{Q}]=\varphi(5) = 4$.
It follows by $(4,5)=1$ that $[L:\mathbb{Q}]=20=2^25$.
Let $\sigma: a \mapsto \xi a$ and $\tau: \xi \mapsto \xi^2$. $N=<\sigma>$ and $K=<\tau>$.
By Sylow's Theorem, $N$ is normal. $N\cap K = e$ hence galois group is given by semidirect product $N \rtimes K$.
Question How can I determine all normal subgroups?
$N$(group of order $5$) is normal.
Since $\tau\sigma\tau^{-1} = \sigma^2$, group is not abelian. Therefore, $K$[$\cong C_4$] (group of order $2^2$) cannot be unique.
How can I argue about subgroups of order $2$?
If subgroup of order $10$ exists, it has index $2$ hence necessarily normal. Is it trivial that it exists?
If $H$ is a normal subgroup of $G = \operatorname{Gal}(L/\mathbb{Q})$, then since $|N|$ is prime, $H \cap N = N$ or $H \cap N = \{e\}$.
If $H \cap N = \{e\}$, let $g = \sigma^m \tau^n$ be an arbitrary element of $H$. Since $H$ is normal, it contains $(\sigma^{-1} g \sigma) g^{-1} = \sigma^{2^n-1} \in N$ (using the relation $\tau^n \sigma = \sigma^{2^n} \tau^n$). Since $H \cap N = \{e\}$ this implies $\sigma^{2^n - 1} = e$, and therefore $4 \mid n$. So $g = \sigma^m \in H \cap N$ and $H$ is trivial.
So a normal subgroup of $G$ is either trivial or contains $N$. The normal subgroups containing $N$ correspond 1-1 with normal subgroups of $G/N \cong K$, which are easy to describe since $K$ is cyclic.