 Solution

For $n=p$ prime, $[K:\mathbb{Q}]=p-1$ and the cyclic group $\mathrm{Gal}(F/K)$ has order either $1$ or $p$. ThereforeThese correspond respectively to $a$ being or not being a $p$th power in $K$. In fact it is a $p$th power in $K$ precisely when it is a $p$th power in $\mathbb{Q}$. To see the nontrivial implication, suppose $\alpha \in K$ and ${\alpha }^p=a$. Taking the field norm gives $b^p=a^{p-1}$ with $b=N_{K/\mathbb{Q}}(\alpha )\in {\mathbb{Q}}^{\times }$. Then $(a/b{)}^p=a$, so $a$ is already a rational $p$th power.

Both degrees occur: $a=1$ gives $p-1$, while $a=2$ is not a rational $p$th power and gives $p(p-1)$. This includes $p=2$, with possible degrees $1$ and $2$.