> Not all decision problems are like that (e.g. happy net problem)
Assuming you have an oracle for any question that you know how to verify, this is very easy to reduce:
Is there a solution in which vertices {1, 4, 6} are all positive?
First, let's establish that the question is legal: presented with a working solution, we can calculate the defining constraint of the problem, and we can check the polarity of vertices 1, 4, and 6. If this problem is in NP, verifying the constraint will take at most polynomial time. If not, not, but our question can be verified in however much time it takes to calculate the constraint.
The complexity of determining a solution by getting answers to questions of this form is interesting.
Case 1: The answer is "no" for all single-vertex sets. All vertices must be negative. This cannot actually happen, because all vertices being negative is the same thing as all vertices being positive.
Case 2: All vertices may be simultaneously positive. In this case, the answer to every question will be "yes", and we'll ask one question for every vertex in the input graph, solving the problem in sublinear time.
Case 3: We need some positive and some negative vertices. By asking about single-vertex sets, we can quickly identify a vertex that can be positive, vertex A.
We will include that vertex in every subsequent question. By the time we get there, we will have already asked about zero or more other vertices and been answered "no". We need never ask about those vertices again; they have to be negative and if we include them in any set, we'll get another "no".
So, let's ask about a never-before-examined vertex, vertex B. Can {A, B} be simultaneously positive?
If not, we can toss B into the "must be negative" pile, since we know A can be positive.
If so, we can include it in our developing solution; future questions will include vertices A and B. We still never need to reexamine any vertex that has ever been included in any of our questions.
So, unless I'm missing something, in this case we'll also produce a solution using one question per vertex in the graph.
> but for this one I don’t see the point making the distinction.
The distinction between NP-complete and NP-hard has nothing to do with the distinction between yes/no questions and open-ended questions. Those are unrelated concepts.
An NP-hard problem is at least as hard as any NP problem. It might be much harder.
An NP-complete problem is NP-hard, but it's also guaranteed to be an NP problem. That's the distinction.