Suppose you find a reduction from the $k$-vertex-cut problem to the hamiltonian-path problem. In particular, you find an algorithm $A$ that, given the graph $G$ and the number $k$, outputs a graph $G'$ such that $G'$ has a hamiltonian path if and only if $G$ can be disconnected by removing $k$ vertices. If $G$ has $n$ vertices, then $A$ runs in $O(n^3)$ time.
If $A$ does work as advertised, does this show that $k$-vertex-cover is $\operatorname{NP}$-hard? $\operatorname{NP}$-complete? In $\operatorname{NP}$?
Since hamiltonian-path is $\operatorname{NP}$-hard, $\operatorname{NP}$-complete, and in $\operatorname{NP}$, I believe the answer is "yes" for all of these.
Suppose $B$ is some problem in $\operatorname{NP}$. If someone discovers $B\in \operatorname{P}$, does this imply that $\operatorname{P}=\operatorname{NP}$?
I think the answer is "no" since all problems in $\operatorname{P}$ are also in $\operatorname{NP}$.
If $B\le_p HAMILTONIAN-PATH$, does this imply that $B\in \operatorname{P}$? $B\in \operatorname{NP}$? That $B$ is $\operatorname{NP}$-complete?
It doesn't imply that $B \in \operatorname{P}$, but it does imply that $B\in \operatorname{NP}$ and that $B$ is $\operatorname{NP}$-complete.
Suppose that someone invents an algorithm to decide if a boolean expression involving variables and logical operations ($\wedge, \vee, > \neg$) could be true, under some assignment of truth values to the variables. If the expression has length $n$, the algorithm runs in $O(n^{12})$ time. Does this imply that $3SAT\in \operatorname{P}$? Does it imply that $\operatorname{P}=\operatorname{NP}$?
I think that this does imply that $3SAT \in \operatorname{P}$, but I'm not sure how to show it. If this is the case, then $\operatorname{P} = \operatorname{NP}$.
Any help on these would be greatly appreciated.
If every instance of problem $A$ is polynomial-time reducible to some instance of problem $B$, that means that problem $B$ is more general (and thus harder) than problem $A$. This is the intuitive idea between polynomial-time reductions and NP-complete proofs. As such, reducing a given problem to a known NP-hard problem does not show that the original problem was NP-hard; it only shows that it is at most as hard as that NP-hard problem, which does not really say much.
In order to prove that a problem is NP-hard, you need to show that a known NP-hard problem is a special case of that problem, i.e., reduce a known NP-hard problem to your problem in polynomial time.
Therefore the answer to the first question (regarding NP-completeness and NP-hardness) is no, since the algorithm $A$ does the converse of what it needs to do; it reduces an arbitrary instance of the $k$-vertex cut problem to some instance of the Hamiltonian path problem, which is $NP-complete$. However, this shows that $k$-vertex cut is in NP, since Hamiltonian path is in NP.
The answer to second question is obviously no, for the reason you mentioned.
The third question is essentially identical to the first one, so it has the same answer. Of course, it also does not imply $B \in P$ since it has nothing to do with this.
The fourth question essentially translates to finding a polynomial-time algorithm that solves the SAT problem, which would imply SAT$\in$P by definition. Further, because SAT is NP-complete all NP problems are polynomial-time reducible to SAT, which means all problems in NP (including 3SAT) would be in P, which, in turn, would mean P=NP.