Is it always possible to have one part of the reduction?
If you define "reduction" using a single implication as in $x\in A \implies f(x) \in B$, then $A$ indeed always "reduces" to any nonempty $B$. Take any $b\in B$, and define $f(x)=b$ the be a constant function. This is computable in O(1), hence in polytime.

It does not matter if $B$ is decidable, or in P, or anything else -- as long as it is not empty, that's enough.

1) No, it does not prove anything. If you have that $A$ reduces to a PTIME $B$ with your proposed reduction, you can't conclude $A$ is PTIME. $HALT$ would reduce to checking if a number is $12$ (which is PTIME), but $HALT$ is surely not PTIME.

2) $f$ above is constant, hence always decidable. (There is no such a thing as a semicomputable function)

3) Similarly, the converse implication is satisfied, for any $A$ and non-full $B$, if we take the constant function $f(x)=b \notin B$.



Below, I consider the more standard definition of reduction with the double implication.

The halting problem is not reducible (polytime or not) to its complement, nor its complement is reducible to the halting problem.

This is because $A\leq B$ holds iff $\bar A \leq \bar B$, hence in either case the complement of $HALT$ would reduce to $HALT$, but the first is not recursively enumerable, and the second is, so we get a contradiction.

For the others:

1) Yes. The reverse direction is trivial, since if $B\in P$ then $B\in NP$ hence it reduces to 3SAT.

2) The $HALT$ counterexample above holds for any computable $f$. There is no such a thing as a semicomputable function -- semidecidability only applies to problems.