Question 1: Why propositional logic has no axioms?
There are different proof systems for propositional calculus; some - called Hilbert-style - have axioms and rules; some, like e.g. Natural Deduction rules only.
When we speak of propositional logic, we usually speak of the language and the calculus: thus, we say that propositional logic is consistent because we cannot derive $\bot$ in the calculus.
Question 2: Why propositional logic is consistent?
We associate to the logic a semantics: for classical logic, the usual semantics is defined via the truth tables for the connectives.
We define the concepts of tautology and tautological consequence as well as the related properties of soundness and completeness.
Soundness implies that the calculus derives only tautologies (while completeness means that the calculus derives all the tautologies).
The key points of soundness are:
Having said that, we immediately have the consistency of the claculus:
$\bot$ is not a tautology.
For an introduction, we can see: S.Simpson, Mathematical Logic (2013).
Doing a truth table is a fine approach. There are four possibilities. For each of them you can see whether the statement each made is correct, then see if that is consistent with the assumption. Here if we assume both A and B told the truth, the statement A made is false, so we have a contradiction. You can keep going in this vein.
Another way is just to assume the truth of one statement and see where that leads. If we assume A tells the truth, we know that B is lying. If B is lying, his statement is false, so we have found a consistent truth assignment. If you trust the puzzle setter to make sure there is only one solution, you can quit here. Otherwise, let us assume A is lying. Then B is lying as well, but with A lying B's statement is true and we have an inconsistency. Therefore A is telling the truth and B is lying.
Best Answer
You can abbreviate the statements as
$M$: multi-user state
$N$: operating normally
$K$: kernel functioning
$I$: in interrupt mode
so that your statments are $$M \iff N, \quad N \implies K,\quad \neg K \vee I, \quad \neg M \implies I, \quad I.$$ You want truth values so that all these statments are true. Clearly $I$ must be true, and in this case so are the third and fourth statements. Just choose values for $M$, $N$, and $K$ that make the first two statements true. One possibility is if $M$, $N$, and $K$ all share the same truth value.