I assume you can already prove that the union, intersection, and complement of regular languages is regular.
Let $L_1$ be regular and $L_2$ be non-regular. Then you should also be able to prove:
At most one of $L_1 \cup L_2$ and $L_1 \cap L_2$ is regular.
If $L_1 \subset L_2$ then their union is irregular. (You should also be able to construct an example).
If $L_2 \subset L_1$ then their union is regular. (You should also be able to construct an example).
If the complement of $L_1$ contains at least two words, there is a regular language $L_3$ such that $L_1 \cap L_3 = \emptyset$ and $L_1 \cup L_3 \subsetneq \Sigma^*$. Then either $L_3 \cup L_2$ is regular or $L_1$, $L_3 \cup L_2$ are an example of a regular and an irregular language whose union is regular but where neither is a subset of the other and their union is not $\Sigma^*$.