= Solution
Put $c=f(1)\geq0$. For a given $x$, let $M=\|x\|^2=\|xx^*\|$ by the <C-star identity>. For $\varepsilon>0$ the binomial series
$$
b_\varepsilon=\sqrt{M+\varepsilon}\left(1-\frac{xx^*}{M+\varepsilon}\right)^{1/2}
$$
converges absolutely in the <Banach algebra>, because $\|xx^*/(M+\varepsilon)\|<1$. Its coefficients are real and $xx^*$ is Hermitian, so $b_\varepsilon^*=b_\varepsilon$; multiplication of the absolutely convergent series gives $b_\varepsilon^*b_\varepsilon=(M+\varepsilon)1-xx^*$. Positivity yields $f(xx^*)\leq(M+\varepsilon)c$. Let $\varepsilon\downarrow0$. Part (iii) now gives
$$
|f(x)|^2\leq c^2\|x\|^2.
$$
Thus $f$ is a <bounded linear functional> with $\|f\|\leq c$. Since the nonzero identity in a <C-star algebra> has norm one, evaluating at it gives the reverse bound. Hence
$$
\boxed{\|f\|=f(1).}
$$
This proves automatic continuity from the original algebraic positivity assumption, including the case $c=0$, when $f=0$.
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