If a problem S is NP-Complete and we know that a problem Q is polynomial time reducible to S. Does that mean that Q belongs to NP?
Also, when can we state that Q is NP-Hard but does not belong to NP?
If S is polynomial time reducible to R, does that mean R is also NP-Complete? (We do not know if R belongs to NP)
If a problem S is NP-Complete and we know that a problem Q is polynomial time reducible to S. Does that mean that Q belongs to NP?
Yes. "$S$ is NP-complete" means two things: $S$ is NP, and $S$ is NP-hard; the second one is irrelevant here, only the first one is needed. Any problem polynomial-time reducible to an NP problem is NP (it works to build the algorithm that first runs the reduction, then runs the non-deterministic algorithm that solves the NP problem).
Also, when can we state that Q is NP-Hard but does not belong to NP?
It suffices to take for $Q$ a problem that is "really high" in the complexity hierarchy. For example, take $Q$ to be your favorite NEXPTIME-complete problem. By the time hierarchy theorem, we know that NP ⊊ NEXPTIME, so if a problem is NEXPTIME-hard, it cannot be NP (otherwise all NEXPTIME problems could be solved in NP by first reducing to this problem in polytime and then applying the NP algorithm, yielding the absurd NP = NEXPTIME).
If S is polynomial time reducible to R, does that mean R is also NP-Complete? (We do not know if R belongs to NP)
No. It only yields that R is NP-hard, but R may not be in NP. For example, SAT is polytime-reducible to any NEXPTIME-complete problem (because SAT is in NEXPTIME), but such problems are not in NP.
