L3 can be proven to be not regular by pumping lemma. Take $ a^{k}b^{k}a^{k}b^{k} $ where k is the pumping length.
L2 is regular because $a^{z z’} \forall ~ zz’ > 0 $ can be rewritten as $ a^{kz’} ~ \forall ~~ k > 0~ for ~a~ given ~z’ $ which is obviously regular. The DFA will have a z’ states and the last state will loop to the first.
L1 can be proven to be not regular by pumping lemma too. Let the pumping distance be $k$. Take the first values of $ a^{z’^{~m}} $ where $ z’ ^ {~m} > k $ . Now $ a^{z’^{~m+1}} = a^ { z’ ^ {~m} \times z’ } >= 2* a^ { z’ ^ {~m} } \forall z’ >= 2$. (Note that when z’=1, the language is regular and the language contains only one string). Now between $ z’^{~m} $ and $2*z’^{~m}$, there lies at least one values of $ k + n*p $ where $ 0<p<=k $ and $ n $ is the number of times the string of length $p$ is pumped. Note that we have taken $ z’ ^ {~m} > k $ . By division algorithm, $z’^{m} = q’ p + r , 0<=r<p$. Now $ z’^{~m}<(q’+1)*p<2*z’^{~m} $. This is not a part of $ a^{z’^{~z}} $. Hence it contradicts pumping lemma.