I know that all algebraic extensions of fields of characteristic $0$ are separable, but what about a field of characteristic $p$, for example, $\mathbb{F}_7$?
I know that, for a finite field of characteristic $p$, all finite extensions are separable, but what happens with infinite algebraic extensions? What if the field itself is infinite?
What you are actually asking about is perfect fields. A field $K$ is called perfect if either the characteristic of $K$ is zero, or the characteristic is a prime $p >0$ and the Frobenius-map $F:K \to K$ associated to $K$, that sends $K \ni \alpha \mapsto \alpha^p$ is surjective (a field homomorphism $K \to K$ is always injective, so actually $F$ is an automorphism of $K$).
Using this definition, you can prove that a field $K$ is perfect iff every algebraic extension of $K$ is separable (over $K$). As you already state, for a finite field $\mathbf{F}_{p^n}$ of characteristic $p>0$ (prime) with $n \in \mathbf{Z}_{>0}$, all finite extensions will be separable. Fortunately, the Frobenius-map associated to $\mathbf{F}_{p^n}$ (sends $\alpha \mapsto \alpha^p$) is an automorphism of $\mathbf{F}_{p^n}$. Therefore, $\mathbf{F}_{p^n}$ is perfect.
However, if you look at the field $\mathbf{F_p}(T)$ where $T$ is a variable, then this variable is not a $p$-th power. Therefore, this is an imperfect field.