1. Yes, it does.  It implies that P != NP, which in turn implies
that no NP-complete problem has a polynomial-time algorithm.  For instance,
there is no polynomial-time algorithm for 3-coloring (or 3SAT, or
the Travelling Salesman Problem, or ...).

2. No (not unless P = NP).  For instance, HornSAT reduces to 3SAT,
but we know of no reduction from 3SAT to HornSAT (and if P != NP, then
no such reduction can exist).  The reason is that HornSAT is in P (it
can be solved in polynomial time), but 3SAT is NP-complete.

To put it another way, if it were true that "reduces to" was a
symmetrical relation, then this would mean that 3SAT reduces to
HornSAT (since it is certainly true that HornSAT reduces to 3SAT).
But since HornSAT is in P, if it were true that 3SAT reduces to
HornSAT, this would mean that 3SAT is in P.  And since 3SAT is
NP-complete, if it were true that 3SAT is in P, it would follow
that P = NP.  Since we don't know of any proof that P = NP,
"B reduces to A" doesn't necessarily follow from "A reduces to B".

Comment: If the question had promised you that both A and B were
NP-complete, then the answer would be "Yes".  But the question didn't
promise you that, so "Yes" is not correct.  (If you wrote "Yes, if
both A and B are NP-complete", that should be worth at least partial
credit.  If you wrote "Yes, if both A and B are NP-complete; no, in
general", that is worth full credit.)