Aussume P != NP, are these assertions valid?
Regarding 1: yes, since $B$ is $\mathsf{NP}$-complete it must be $\mathsf{NP}$-hard, i.e., for ever every problem $C$ in $\mathsf{NP}$ we have $C \le_P B$. Since $A \in \mathsf{P} \subseteq \mathsf{NP}$, we can pick $C=A$.
This is true regardless of whether $\mathsf{P}=\mathsf{NP}$.
Notice, however, that we are using the fact that $B$ is $\mathsf{NP}$-complete. In general it is not possible to reduce an arbitrary problem in $\mathsf{P}$ to an arbitrary problem in $\mathsf{NP}$.
Regarding 2: no. Here your reasoning is correct, $B \le_p A$ would imply $\mathsf{P} = \mathsf{NP}$.