Basic: A language is not regular Language if you need an extra Data structure for keeping track of alphabet.
Here the same problem:
In L1, we need to keep track of 'a' so that you can compare with a number of 'b'. So L1 is not regular Language.
Similarly, in L2 = { a^nb^n, a^nb^na^nb^n, ............}. here we also need to compare a number of 'a' and 'b'. So, L2 is also not a regular language.
If anything not clear, please comment, I try to simplify more.