The Galois group for any cyclotomic field $\mathbb{Q}(\zeta_n)$ will be the multiplicative group $\mathbb{Z}_n^{\times}$. When $n$ is prime, as is the case here, then $\mathbb{Z}_n^{\times} \cong \mathbb{Z}_{n-1}$, which is cyclic. (The latter fact can be proven using the structure theorem for finitely generated abelian groups.)
Therefore, we simply need to find a generator for the Galois group. To do this, note that every $\mathbb{Q}$-automorphism of this field is determined by its action on $\zeta$, and further, $\zeta \mapsto \zeta^k$ always defines a legitimate automorphism (since $\text{Gal}(f)$ acts transitively on the roots of $f$ when $f$ is irreducible). Therefore, I simply experiment as follows:
Does $\phi$ such that $\phi(\zeta) = \zeta^2$ generate the group? If not, what about $\phi(\zeta) = \zeta^3$? And I continue until I have found my generator -
it'll end up being $\zeta \mapsto \zeta^k$ for a $k$ that generates $\mathbb{Z}_n^\times$.
Once you've found the generator $\phi$, the Galois group is simply $\{ \text{id}, \phi$, $\phi^2$, $\phi^3\}$. Noting that $\mathbb{Z}_4$ has one subgroup isomorphic to $\mathbb{Z}_2$, you will get a subgroup $\{ \text{id}, \phi^2\}$ that sends $\zeta \mapsto \zeta^4$ and $\zeta^2 \mapsto \zeta^3$ when it acts on the roots.
Long story short, I think this is simpler than trying to think about the automorphisms in terms of complex conjugation, etc.
As you already know, the elements of the Galois group are exactly: $\newcommand{\Q}{\mathbb{Q}} w \mapsto w, w \mapsto w^2, w \mapsto w^4, w \mapsto w^5, w \mapsto w^7, w \mapsto w^8$. $w \mapsto w^2$ is a generator, and we have
$$
w \mapsto w^2 \mapsto w^4 \mapsto w^8 \mapsto w^7 \mapsto w^5 \mapsto w.
$$
Thus we are looking for the fixed fields of $w \mapsto w^4$ and of $w \mapsto w^8 = w^{-1}$.
The fixed field of $\boldsymbol{\sigma: w \mapsto w^{-1}}$
As you have observed, $\alpha = w + w^{-1}$ is fixed by $\sigma$ and satisfies
$$
\alpha^3
= w^3 + w^{-3} + w^1 + w^{-1}
= (w^6 + 1)/(w^3) + \alpha
= \alpha - 1
$$
Since $\alpha^3 - \alpha + 1$ is irreducible over $\Q$, the desired fixed field is $\Q(\alpha) = \Q(w + w^{-1})$.
The fixed field of $\boldsymbol{\tau: w \mapsto w^4}$
$\beta = w^3$ satisfies $\beta^2 + \beta + 1$, which is irreducible over $\Q$ of degree $2$. So the desired fixed field is $\Q(\beta) = \Q(w^3)$.
A Remark
The generators for a subfield can end up being really simple: in our case $w + w^{-1}$ and $w^3$. But how do you find them?
The first thing to do is try looking for a single term or two terms added together that is fixed under the automorphism in question. In our case, $w^3$ is a single term so you would have found it quickly; $w + w^{-1}$ would not take much longer.
Alternatively, expecially if the first method fails, you can write an arbitrary element $z \in \Q(w) / \Q$ as $z = a w^5 + b w^4 + c w^3 + d w^2 + e w + f$, manually set $\sigma(z) = z$ and then solve for equations in $a, b, c, d, e, f$. You will use the minimal polynomial for $w$ (or in general, the minimal polynomial for a generator of whatever extension you are considering) to simplify powers of $w$ above $w^5$, and then equate each coefficient $w^0, w^1, \ldots, w^5$. This will give you a set of things fixed by the autorphism, and picking one such list of $a, b, c, d, e, f$ chances are good you will get a generator of the desired field.
A Second Remark
This method should always work, though it could potentially be more tedious.
Say we have an extension $F(\alpha) / F$ of degree $n$ with Galois group $G$.
First, enumerate all the automorphisms of the extension by casework on where they send $\alpha$, and use them
to classify the Galois group $G$.
Then enumerate the subgroups, and for each subgroup $H$, look for elements $z \in F(\alpha)$
that are fixed by $H$.
Once you have an element $z$ with $[F(z) : F] = [G : H] = n / |H|$, you know that
$z$ generates the corresponding extension.
This works for any $H$; $H$ need not be normal in $G$.
For a Galois extension $K / F$ with group $G$ the size of a subgroup $H$
always equals the degree $[K : E]$, where $E$ is the fixed field of $H$.
In other words, the number of automorphisms of $K / F$ fixing $E$ equals the degree of $K / E$.
Equivalently, $\boldsymbol{K/E}$ is Galois for any $H$.
What normality of $H$ corresponds to is that $\boldsymbol{E / F}$ is Galois, i.e. that the number of automorphisms
of $E / F$ fixing $F$ equals the degree $[E : F]$.
In this case the Galois group of $E / F$ is the quotient group $G / H$.
A good example is the extension $K / \Q$, where $K$ is the splitting field of $x^3 - 2$.
If $\omega$ is a cube root of unity, the subextensions are generated by
$\omega, \sqrt[3]{2}$, $\omega \sqrt[3]{2}$, and $\omega^2 \sqrt[3]{2}$,
and have degree 2, 3, 3, and 3, respectively over $\Q$.
The Galois group is $S_3$ and the corresponding subgroups are
of index 2 (normal), index 3 (not normal), index 3 (not normal), and index 3 (not normal), respectively.
So the index of the subgroup always equals the degree of the extension.
Normality doesn't factor in unless we consider the automorphism group of a subextension rather than
the subgroup fixing it.
Best Answer
Highlights
Put $\;\zeta:=e^{\frac{2\pi i}5}$, and let $\;\omega\in Gal(\Bbb Q(\zeta)/\Bbb Q)\;$ be complex conjugation.
Observe that for$\;z\in\Bbb C\;\;,\;\;\overline z=z^{-1}\iff |z|=1\;$ , and thus:
$$\omega(\zeta+\zeta^{-1})=\omega(\zeta)+\omega(\zeta)^{-1}=\zeta^{-1}+\zeta\implies \zeta+\zeta^{-1}\in\Bbb R$$
(or, of course, simpler: $\;\zeta+\zeta^{-1}=\zeta+\overline\zeta=2\,\text{Re}\,\zeta\in\Bbb R)\;$
which means
$$\zeta+\zeta^{-1}\in\Bbb Q^{\langle\omega\rangle}:=\;\text{the fixed field of}\;\;\omega$$
and since the order of $\;\omega\;$ in the Galois group is two and clearly $\;\zeta+\zeta^{-1}\notin\Bbb Q\;$ , we're done.
You can also check try to find the irreducible rational polynomial of $\;\zeta+\zeta^{-1}\;$ ...and it is a quadratic.
Finally, just check that
$$2\cos\frac{2\pi}5=\frac12(\sqrt5-1)\;\ldots$$