Prove, axiomatically that $1$ does not equal $0$.

If you start with Peano axioms for the natural numbers, then $0$ is part of the language, but $1$ is not. We use $1$ as a shorthand for the term $s0$.

Now we can use the axiom $\forall x(sx\neq0)$, and infer that in particular for $x=0$ it is true that $s0\neq0$. Congratulations, we proved that $0\neq1$ axiomatically.

You can choose different contexts, like set theory, field theory, ring theory or other contexts in which we can interpret $0$ and $1$. You can also find contexts in which $0=1$ is a provable statement. For example the theory whose single axiom states $0=1$. True this theory describes very little of what we expect from the natural numbers, or the symbols $0,1$ to mean. But it is a mathematically valid thing to do.

1) Prove $a \times 0 = 0$ for all $a$.

$a \times 0 = a(1 - 1) = a - a = 0$.

2) If $0 = 1$ then $a = a \times 1 = a \times 0 = 0$

So all terms equal 0.

Which isn't actually a contradiction. It just means we are working with a trivial field. If the field isn't trivial (say the Reals) than $0 \ne 1$.

Maybe we have an axiom that says $0$ is the additive identity and another axiom that says $1$ is the multiplicative identity, that is to say, $x + 0 = x$ and $x \times 1 = x$. Then, in order for $0 = 1$, we'd need to have $1 + 1 = 1$ and $0 \times 0 = 0$. But this implies that $1 \times 2 = 1$, contradicting the multiplicative identity axiom which requires $2 \times 1 = 2$. (Maybe we also need an axiom saying addition and multiplication are commutative).