Completeness of a quantifier-free axiomatization of Boolean algebra using partial-ordering

Axiom 8 can be replaced by the axiom $a\vee(b\wedge c)=(a\vee b)\wedge (a\vee c)$. In the language of order theory, axioms 1-6 describe a (bounded) lattice, and adding axiom 7 gives you a complemented lattice (with a specific choice of complement operation). Boolean algebras are the same thing as complemented distributive lattices, so all you need to add to axioms 1-7 is the distributive law.

Here's an answer to the first question: $(8)$ cannot be proved from $(1)$-$(7)$. There is in fact a nice countermodel, namely the lattice $M_3$ equipped with a "negation" operation which yields a derangement of the intermediate elements.

More precisely, our countermodel will have underlying set $\{0,1,x,y,z\}$. The partial order is given by setting $0$ at the botom, $1$ at the top, and no other inequalities holding besides reflexivity; $\wedge$ and $\vee$ are the the usual join and meet operations. See here for a picture of this. We then let negation be the unary function sending $0$ to $1$, $1$ to $0$, and $$x\mapsto y\mapsto z\mapsto x.$$

Incidentally, the lattice $M_3$ itself (so, the countermodel without negation) is an important example in order theory; it's one of the two lattices characterizing non-distributivity, the other being $N_5$.

Of course, this leaves the second question open. EDIT: per Eric Wofsey's answer, the answer to the second question is yes.