When is $\overline{K}/K$ a Galois extension of $K$?

The argument for the first two is correct. The third field is neither charateristic $0$ nor finite.

In fact, it turns out in the last case the extension is not separable, and the extension not Galois. To see this consider the polynomial $X^p -t$. It is irreducible over $F_p(t)$ yet is not separable.

As others have said, your reasoning is correct for the first two examples but not the third. Let me discuss how to answer the question for a general field.

In general, if $K$ is a field, then $\overline{K}/K$ is Galois iff either $K$ has characteristic $0$ or $K$ has characteristic $p$ and every element of $K$ has a $p$th root. Such a field is called perfect. Indeed, if some element $a\in K$ does not have a $p$th root, then $\sqrt[p]{a}\in \overline{K}$ is not separable over $K$. Conversely, if every element of $K$ has a $p$th root and $a\in\overline{K}$ is not separable, let $f(x)$ be the minimal polynomial of $a$. Since $a$ is not separable, we can write $f(x)=g(x^p)$ for some polynomial $g(x)\in K[x]$. But since every element of $K$ has a $p$th root, we can take the $p$th roots of all the coefficients of $g$ to get $h(x)\in K[x]$ such that $f(x)=h(x)^p$. This contradicts irreducibility of $f(x)$.

In particular, a finite field is perfect since $x\mapsto x^p$ is injective and hence surjective since any injection from a finite set to itself is surjective. But $\mathbb{F}_p(t)$ is not perfect, since $t$ has no $p$th root.