The field of rational numbers $\Bbb Q$ is usually the first field one encounters, and so extensions of $\Bbb Q$ like $\Bbb Q(i)$ or $\Bbb Q(\sqrt 2)$ are the first manifestations of Galois groups one encounters.
Even then, the next basic fields are the finite fields $\Bbb F_p$ (of which $Gal_{\Bbb F_p}(\overline{\Bbb F_p} )$ is well understood), and then local fields like $\Bbb Q_p$ (whose absolute Galois group is a bit more complicated, but still "simple").
While not much can happen in finite or local fields, the Galois theory of global fields ($\Bbb Q, \Bbb F_p(X), \Bbb C(X), \ldots$) is much richer.
When you being studying field extensions of $\Bbb Q$ without Galois theory, there are "hard" questions like "okay, if I adjoin a root of $-1$ or a root of $2$ I get a degree $2$ extension, but what about adding both at once ? How do I know that $\sqrt 2 \notin \Bbb Q(i)$ or vice versa ?", or even harder, "how can I know that $\sqrt {11} \notin \Bbb Q(\sqrt 2,\sqrt 3, \sqrt 5,\sqrt 7)$ ?" (you can find those kind of questions on this website).
The general answer to these questions rely on understanding the Galois group of $\Bbb Q(\sqrt{-1},\sqrt 2,\sqrt 3,\sqrt 5,\ldots)$, and you can translate this in terms of the absolute Galois group by giving an explicit description of $\hom (Gal_\Bbb Q(\overline {\Bbb Q}) , \Bbb Z/2\Bbb Z)$.
A very related kind of question is what happens locally in extensions of global fields, or "what primes factors in $\Bbb Q(\sqrt 7)$ ?" and more generally "how do primes of $K$ factor in primes of $L$ in an algebraic extension $K \subset L$ ?". For our quadratic extensions, this is summed up by Gauss' quadratic reciprocity law, (a result that seemed unbelievable when I first heard about it).
A theorem of Kronecker says that every abelian extension of $\Bbb Q$ is in a cyclotomic extension. If we understand cyclotomic extensions (and we do) then we understand morphisms from $G$ to finite abelian groups, which means we understand the abelianization of $G$. To put it concisely $\Bbb Q \subset \Bbb Q(\zeta_n)$ has Galois group $(\Bbb Z/n\Bbb Z)^*$, and we know how a prime $(p)$ behaves in the extension by looking at what $p$ (we pick the positive generator) modulo $n$ does in that group.
A big achievement of the 20th century was to generalize this result to any global field, notably number fields other than $\Bbb Q$, so we have a "nice" description of $G^{ab}$ and a reciprocity map $\{primes\}\to G^{ab}$.
With all of this we can answer new questions like "But how do I know that $\sqrt[3]{11} \notin \Bbb Q(\sqrt[3]2, \sqrt[3]3,\sqrt[3]5,\sqrt[3]7) ?"$ or "when is $2$ a cube modulo $p$ ?" because this is about abelian extensions of $\Bbb Q(\sqrt{-3})$
So far we understand pretty well the morphisms (representations) from the absolute Galois groups to finite abelian groups. The question about representations into more complicated groups is an active area of research and an answer about those would give us tools to answer easily more of those kind of "simple" questions.
For example if $f(a)$ is the positive root of $x^5-x-a$, I don't think we have an easy answer to "But how do I show that $f(5) \notin \Bbb Q(f(1),f(2),f(3),f(4))$ ?"
Best Answer
As said in the comments by Sofie Verbeek, if $F/\Bbb Q \to Gal(F/\Bbb Q)$ was representable it would be a functor on the form $F/\Bbb Q \mapsto Hom(F, F_0)$ for some fixed field $F_0$.
We assume that such $F_0$ exists, let $F$ be a field with bigger cardinality than $F_0$, we see now that $Gal(F/\Bbb Q) = \emptyset$, a contradiction.
Tautologically, the functor $F \to Hom(F, \overline{\Bbb Q})$ represents the functor $F \mapsto \{\text{embedding $\sigma : F \to \overline{\Bbb Q}$}\}$.
Edit : This answer is wrong but I can't delete it because it was accepted by the OP. So please moderators if you see this answer I would appreciate if you can delete it.