Prove/disprove: If $_1$ is a finite language but not empty and $_2$ is NOT regular then $_1 \circ _2$ is NOT regular
$L_1\circ L_2$ can certainly still be non-regular. For example, when $L_1$ contains exactly one word.

However, $L_1\circ L_2$ can be regular, too. Here is an example. Let $L_1=\{\epsilon,0\}$. Let $L_2$ be the complement of $\{ 0^n1^n \mid n \gt 0 \}$.
As the complement of a non-regular language, $L_2$ is not regular while $L_1\circ L_2$, the set of all words over $\{0,1\}$, is regular.



Here are three similar exercises.

Exercise 1. Construct examples when the alphabet contains exactly one letter. 


Exercise 2. Prove or disprove: If $L_1$ is a finite language but not empty and $L_2$ is not context-free, then $L_1\circ L_2$ is not context-free.



Exercise 3. Prove or disprove: If $L_1$ is a finite language but not empty and $L_2$ is not recursive, then $L_1\circ L_2$ is not recursive.