Can the difference of a non-regular and a regular language be regular?
When $L_2 = \Sigma^*$, then $L_3 = \emptyset$ no matter what $L_1$ is.



Let us say that a language $L$ is subset-regular if $L \cap L'$ is regular for all languages $L'$. In other words, $L$ is subset-regular if all of its subsets (including $L$ itself) are regular.

Theorem. A language is subset-regular iff it is finite.

Proof. Clearly every finite language is subset-regular. In the other direction, an infinite language has uncountably many subsets, so not all of them can be regular. $\quad\square$

We can replace $\Sigma^*$ above with the complement of any subset-regular language, that is, with any cofinite language. Moreover, due to the theorem above, only cofinite languages work: if $L$ has a non-regular (proper) subset $L'$, then $\overline{L'} \setminus \overline{L} = \overline{L'}$ isn't regular.