Intersection and union of a regular and a non-regular language
We can prove this by contradiction. 
Lets define $\overline{L_1} = \Sigma^* \setminus L_1$. Then we can reformulate $L_2$:

$L_2 = ((L_1 \cup L_2) \setminus L_1) \cup (L_1 \cap L_2) = ((L_1 \cup L_2) \cap \overline{L_1}) \cup (L_1 \cap L_2)$

We know:


Regular Languages are closed under union, intersection and complement
$\overline{L_1}$ and $L_1 \cap L_2$ are regular
$L_2$ is not regular


Now assume $L_1 \cup L_2$ is regular: Then $((L_1 \cup L_2) \cap \overline{L_1}) \cup (L_1 \cap L_2)$ is regular (as it is only a union/intersection of regular languages), so $L_2$ would be regular. That is a contradiction, therefore our assumption is false, and $L_1 \cup L_2$ can not be regular.