Understanding facts about regular languages, finite sets and subsets of regular languages
One useful fact about regular languages is that a union of finitely many regular languages is regular. From this it follows that all finite languages are regular: the language that recognizes a single constant string is always regular, and a finite language is a union of finitely many constant strings.

(Proof: remember that regular languages are exactly those representable by regular expressions. A regular expression can be trivially written to recognize a single constant string. To recognize a finite union of regular languages, simply combine their regular expressions with the union operator. Now you have a (potentially enormous but) finitely-long regular expression recognizing the entire language.)

This means that 1 is true, because a finite set is equal to the union of all its proper subsets, and an infinite language always has a non-regular subset.

(Proof: if a language $L$ is finite, then it has finitely many subsets, and is equal to the union of these subsets. If $L$ is infinite, then it has uncountably many subsets. But there are only countably many regular languages over a given alphabet, because regular expressions are strings and thus can be enumerated. Thus by diagonalization there exists a non-regular subset.)

2, on the other hand, is false as long as there exists an infinite non-regular language. (And such a language does exist: $\{ 0^n 1^n | n \in \mathbb{N} \}$ is a classic example.) It's not regular, but all its finite subsets are finite, and thus regular.

Your intuition is correct: 3 is false. The non-regular language I just mentioned is a proper subset of $\{ 0^a 1^b | a, b \in \mathbb{N} \}$, which is regular.

4 is also true, since any subset of a finite set is always finite.