= Solution
Set $A_0=x(a^*a)$, $B_0=x(b^*b)$ and $c=x(a^*b)$. The preceding reality argument gives $x(b^*a)=\bar c$. Positivity therefore implies, for every $\lambda\in\mathbb C$,
$$
0\leq x((b+\lambda a)^*(b+\lambda a))=B_0+\lambda\bar c+\bar\lambda c+|\lambda|^2A_0.
$$
If $A_0>0$, choose $\lambda=-c/A_0$ to obtain $0\leq B_0-|c|^2/A_0$. If $A_0=0$ and $c\ne0$, choosing $\lambda=-tc$ and letting $t\to\infty$ contradicts positivity. Thus $c=0$ in that case too. This proves the <Cauchy–Schwarz inequality for positive C-star functionals>:
$$
\boxed{|x(a^*b)|^2\leq x(a^*a)x(b^*b).}
$$
Apply it with $a=1_A$, $b=z$. Since $0\leq z^*z\leq\|z\|^21_A$, positivity gives
$$
|x(z)|^2\leq x(1_A)x(z^*z)\leq x(1_A)^2\|z\|^2.
$$
Thus $\|x\|\leq x(1_A)$, while evaluation on $1_A$ gives the reverse inequality. Consequently
$$
\boxed{\|x\|=x(1_A).}
$$
The zero functional is covered as well. In the trivial zero algebra all assertions are immediate, and its state space is empty.
For a nonzero unital <C-star algebra>, the set $S_A$ of <state on a C-star algebra> functionals is <convex>, since both positivity and evaluation $x(1_A)=1$ are preserved by <convex combinations>. The <norm> identity places it in the closed dual unit ball. Suppose a state satisfies $x=ty+(1-t)z$, with $0<t<1$ and $\|y\|,\|z\|\leq1$. Evaluating at the unit and taking real parts gives
$$
1=t\operatorname{Re}y(1_A)+(1-t)\operatorname{Re}z(1_A)\leq t+(1-t)=1.
$$
Since both weights are strictly positive, both real parts equal $1$. Also $|y(1_A)|,|z(1_A)|\leq1$, so $y(1_A)=z(1_A)=1$. Hence
$$
1=|y(1_A)|\leq\|y\|\leq1,\qquad1=|z(1_A)|\leq\|z\|\leq1.
$$
The stipulated converse to the <norm> identity makes $y,z$ positive, so both belong to $S_A$. Therefore \b[$S_A$ is a <face of a convex set>, namely the closed dual unit ball]. More precisely it is the <exposed face of the dual ball defined by the unit>, since it is cut out by $\operatorname{Re}x(1_A)=1$.
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