If a language is finite, it is definitely regular. However, if a language is infinite, it may or may not be regular. $(0 + 1)^*$ is regular, yet infinite. $0^n1^n$ is also infinite, but non-regular.
If there is a pattern in the language that you can exploit, there likely is a DFA/NFA for it. If the language looks like it is non-regular but actually is finite, it is definitely regular and you can make a DFA/NFA for it.
$L_1$ is non-regular because we have to keep track of how many $a$'s vs $b$'s, or $b$'s vs $c$'s, and the language is infinite as well, so we can't create a DFA for it (finite memory of a DFA not equipped to track the counts, would need a stack).