You can see :
for a complete treatment of the proof system based on the Resolution Rule.
The proof system "expands" the set of cluases, applying the Resoultion Rule in order to add e new clause (a disjunction) to the original set of clauses. The rule has the property that "preserves stisfiability", i.e. the new set of clauses is satisfiable iff the original one is.
Thus, if after iterated applications of the rule the empty clause is produced, being it unsatisfiable (it is another way of writing $\bot$, i.e. a contradiction) this is enough to conclude that also the original set of clauses is unsatisfiable.
Yes. That is exactly what it means. Consistency assumptions are axioms.
This gives rise to a natural hierarchy of axioms, specifically part of set theory, called large cardinal axioms which are stronger and stronger in consistency strength, and generally each one implies the weaker are consistent (and much more).
For example, the standard set theory, ZFC (Zermelo–Fraenkel with Choice) does not prove its own consistency strength, but we can add an axiom stating that it is in fact consistent. To that you can add an axiom that it is not only that ZFC is consistent, but "ZFC+ZFC is consistent" is also consistent. This can go on for a while.
But you can also just say that there are inaccessible cardinals, whatever they might be. This implies that ZFC is consistent, and much more. You can move to stating that there exists a weakly compact cardinal, which then implies that not only it is consistent that there is an inaccessible cardinals, but that it is consistent that every set is smaller in size from some inaccessible cardinal.
And the list continues.
Interestingly, though, while the large cardinal axioms are stating that some particular sets exists (or don't exist), consistency statements can be seen as axioms added to the theory of the natural numbers. So you can also investigate them from arithmetic theories such as PA or PRA, both of which are vastly weaker than ZFC.
Best Answer
The key point is that "proof" requires you to at least understand what theory you are assuming.
Both the Riemann Hypothesis and "$T$ is consistent" (where $T$ is any recursively enumerable theory) are $\Pi^0_1$ statements. This means that if there is a counterexample, then the statement is disprovable. Or in other words, if the statement is not disprovable, then it is true.
But what does "true" mean? In the context of arithmetic true means "true in the natural numbers". In other contexts, it is usually meaningless, or synonymous for "provable".
Yes. $\sf PA$ is consistent is a $\Pi^0_1$ statement, but we also can prove that it is a true statement, since $\Bbb N$ exists and it is a model of $\sf PA$. Or at least that is the case if your metatheory is sufficiently strong (e.g. set theory à la $\sf ZFC$). The point is that $\sf PA$ itself cannot prove that $\sf PA$ is consistent, and that $\sf ZFC$ cannot prove that $\sf ZFC$ is consistent.
The difference between that and the Riemann Hypothesis is that you don't really talk about consistency of a theory. So there is no natural candidate of a theory which is insufficient for proving the Riemann Hypothesis.
It is possible for a $\Pi^0_1$ statement to turn out equivalent to "$\sf ZFC$ is consistent", in which case it cannot be proved. But it would be a mind shattering surprise to modern mathematics if RH turns out to be such statement. So what we expect is that RH is either outright refutable from a weak theory like $\sf PA$, and if not then it is at least true in $\Bbb N$, which is "the thing we care about" for the Riemann Hypothesis.