Prove/disprove: If $_1$ is a finite language but not empty and $_2$ is NOT regular then $_1 \circ _2$ is NOT regular That what I have so far, but I am not sure at all.

Assume toward contradiction that $_1 \circ _2$ is regular.

Define $\Sigma' = \{\sigma'|\sigma\in\Sigma\} $.

Define a regular substitution $\forall\sigma\in\Sigma, h(\sigma) = \sigma'$.

$h(_1) = \{ = '_1'_2 ... '_ |  = _1_2 ... _ \in _1\}$ is also regular and finite by the closure of RL under substitution. 

Therefore, by the assumption $h(_1) \circ _2$ is also regular.
Define regular substitution $\forall\sigma'\in\Sigma',g(\sigma')=\varepsilon,\forall\sigma\in\Sigma,g(\sigma)=\sigma$.

$g(h(L_1) \circ _2) = _2$ which is regular by the closure of RL under substitution, in contradiction to the assumption.
Therefore, $_1 \circ _2$ is not regular. Q.E.D

I am not sure about this statement "$h(_1) \circ _2$ is also regular".

Thanks for any help!!
$L_1\circ L_2$ 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.