deductive system where the axioms are all the propositions that are not tautology

Question

let D be a deductive system. The system's axiom set, A, includes all propositions that are not tautology, and the only rule of inference is $\frac{\alpha \vee \beta}{\alpha \wedge \beta}$.

I need to answer:

1. Show with induction on the proof that $ \vdash \alpha \Longrightarrow \nvDash \alpha $, i.e if $ \alpha $ can be proven from the empty set, then exists a model $ M $ such that for every $ \varphi \in A $, $ M(\varphi)=T $ and $ M(\alpha)=F$.
2. Is the system strongly sound?
3. Is $X\nvDash \alpha \Longrightarrow X\vdash \alpha$?
4. Prove that the empty set isn't consistent in D

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1. I've already asked about this part of the question here but I still have a problem with it, because every axiom, $a\in A$, $\vdash a$ and $A \models a$, in contradiction to what I'm supposed to prove here. Moreover because the axiom set includes all what isn't tautology it also includes all that is contradiction, which means that there is no model $ M $ such that for every $ \varphi \in A $, $ M(\varphi)=T $ which means for every proposition $\psi$, $A\Longrightarrow \psi$.
2. This one was fairly easy base on (1) the system isn't sound so it can't be strongly sound.
3. I'm not even sure how to approach this, I don't even have an intuition whether this is correct or not. Assuming this is wrong, I need to find a set $X$ and proposition $\alpha$ such that $X\nvDash \alpha$ but $\alpha$ cannot be proven from $X$, and if it's correct I need to show that for every set, $X$, and every proposition $\alpha$ such that $X\nvDash \alpha$, $\alpha$ can be proven from $X$
4. On the surface this looks pretty easy, both $(\varphi \wedge \psi)$ and $\neg(\varphi \wedge \psi)$ are axioms, as they aren't tautology, hence $\vdash (\varphi \wedge \psi)$ and $\vdash \neg (\varphi \wedge \psi)$ and the empty set isn't consistent but this seems too easy, like I'm missing something.

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EDIT:

I might have come up with a solution for (3), if $X\nvDash \alpha$ it means that $\alpha$ isn't a tautology and if it isn't a tautology it is an axiom in D and therefore can be proven from $X$