Identify the language

Question

Consider the following languages:
L1={aⁿ bⁿ:n≥0 and n is not multiple of 5}
L2=complement of {(0ⁿ 1ⁿ )^m│m,n>0}

Answer 1

Let L_n = {0ⁿ b ⁿ : n ≥ 0} and L_w = {w ∈ {a, b} ∗ : |w| is not a multiple of 10}.

It is clear that L_nis context-free, and L_w is a regular language.

Furthermore, we have L = L_n ∩ L_w, since the intersection of a context-free language and a regular language is context-free, then L is context-free.
So L₁ is not regular.

Source : https://raphiinconcordia.files.wordpress.com/2015/05/assgn3-sols.pdf

Second Problem:
L₂=complement of {(0ⁿ 1ⁿ )^m│m,n>0}

L₄ = {(0ⁿ 1ⁿ )^m│m,n>0}

It is not possible to construct any DFA ,NFA or RE for L₄

So L₄ is not regular.
Applying Closure property, we say , L₂ is not regular.

Second problem Explanation:  Let L =  {(0n 1n )│n>0}  We know that L is not regular.  Lc = {(0n 1n )m│m,n>0} So we can say Lc is closure of L.  As L is not regular, Lc is not regular.  The complement of  Lc  (L2) is also not regular.

Comments

I know this is not regular.. but how can I be sure about that it is CFL?

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I have edited answer. Just have a look.
I think that the answer would clear your doubt.

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Use closure properties of CFL, you shall get that second language is CFL