a^nb^n is regular, but it is also context free. I don't think there's a restriction on the n. Why do you say this?

Edit: sorry, I read "finite" as "infinite" :0 But n can be infinite and a^nb^n is still regular, and also context free. To be clear, the Chomskky Hierarchy of formal languages goes like this:

Finite ⊆ Regular ⊆ Context-Free ⊆ Context-Sensitive ⊆ Recursively Enumerable

That's because formal languages are identified with the automata that accept them and when an automaton accepts e.g. the Recursively Enumerable languages, then it also accepts the context-sensitive languages, and so on all the way down to the finite languages. One way to think of this is that an automaton is "powerful enough" to recognise the set of strings that make up a language.

Yes, all finite languages are regular.

Specifically, you can construct a finite automata to represent it.