Manipulating Intersection of oracles
The answer to all of those questions is no, and not for some especially deep reason, as most of these can be ruled out by forcing the intersection to be simple enough.

Suppose $B$ contains all languages with only even length words, and $C$ contains all languages with odd length words. Clearly $B\cap C =\emptyset$, so $\mathsf{P^{B\cap C}=P}$, but $P^B, P^C$ both contain undecidable problems. This example rules out $1,2$.

To rule out $3$ you can set $\mathsf{B=EXP}$ and $C=\emptyset$. Choosing $B=\emptyset$ rules out $4$. Not allowing the empty set won't change much, as we can always do the same tricks with finite languages (or simple infinite langauges), so it is probably not possible to claim anything interesting in the general case.