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Solution

For $L_1$:

$L_1=\{a^{2^z}\mid z\text{ is an integer}\}$

The possible lengths are:

$1,2,4,8,16,32,\ldots$

So,

$L_1=\{a,a^2,a^4,a^8,a^{16},\ldots\}$

This is not regular because the gap between consecutive string lengths keeps increasing.

So, $L_1$ is not regular.

For $L_2$:

$L_2=\{a^{2z}\mid z\ge 0\}$

The possible lengths are:

$0,2,4,6,8,\ldots$

So,

$L_2=\{\lambda,a^2,a^4,a^6,\ldots\}$

This language contains all strings of even length over alphabet $\{a\}$.

It can be written as:

$L_2=(aa)^*$

Since $(aa)^*$ is a regular expression, $L_2$ is regular.

For $L_3$:

$L_3=\{\omega\omega\mid \omega\in\{a,b\}^*\}$

In this language, the second half of the string must be exactly the same as the first half.

A finite automaton cannot remember the complete first half and compare it with the second half.

So, $L_3$ is not regular.

Therefore, only $L_2$ is regular.