= Solution
For the converse, let $f$ be bounded and $\|f\|=f(1)=c$. If $c=0$, it is zero and positive. Suppose $c>0$. For any Hermitian $h$, the <Banach algebra exponential> $e^{ith}$ is unitary for real $t$, so its norm is one by the <C-star identity>. Consequently
$$
|f(e^{ith})|\leq c,\qquad f(e^{ith})=c+itf(h)+O(t^2).
$$
Writing $f(h)=u+iv$ and squaring moduli gives $|f(e^{ith})|^2=c^2-2cvt+O(t^2)$. The bound for both signs of small $t$ forces $v=0$, so $f(h)$ is real.
For a Hermitian element $a$, repeated applications of the C-star identity give $\|a^{2^n}\|=\|a\|^{2^n}$. The <spectral radius formula> therefore gives $\|a\|=r(a)$. Take $h=x^*x$; by the permitted spectral positivity assumption, $\sigma(h)\subseteq[0,\|h\|]$. If $M=\|h\|>0$, affine spectral mapping and the just-proved Hermitian norm identity give $\|1-h/M\|\leq1$. Hence $|c-f(h)/M|\leq c$. Since $f(h)$ is real, this implies $f(h)\geq0$. If $M=0$, $h=0$ and the result is immediate. Thus \b[a bounded functional with $\|f\|=f(1)$ is positive].
For the spectral-value request, let $\lambda\in\sigma(x)$ and define a functional on the linear span of $1,x$ by $g(a1+bx)=a+b\lambda$. If $b\ne0$, the scalar $a+b\lambda$ belongs to $\sigma(a1+bx)$: subtracting it from that element gives $b(x-\lambda1)$, which is not invertible. The same norm bound is immediate for $b=0$. Therefore
$$
|a+b\lambda|\leq r(a1+bx)\leq\|a1+bx\|.
$$
This also proves well-definedness if that span is one-dimensional. We have $g(1)=1$ and $\|g\|=1$. The complex <Hahn-Banach theorem> extends it to a functional $f$ on all of $A$ with norm one and $f(1)=1$. The converse just proved makes that extension positive. Thus \b[there is a state with $f(x)=\lambda$], the <spectral values attained by C-star states> result.
Finally, for $x\ne0$, write $x=h+ik$ with $h,k$ Hermitian. At least one of them is nonzero. Its norm equals its spectral radius, so compactness of its nonempty spectrum supplies a nonzero spectral value. Apply the preceding state construction to that Hermitian element. Part (i) makes $f(h),f(k)$ real; the selected one is nonzero, so $f(x)=f(h)+if(k)\ne0$. This proves \b[positive functionals separate every nonzero element], the <states separate elements of a C-star algebra> property. Looking only at $\sigma(x)$ would not prove this last step, since a nonzero quasinilpotent $x$ can have spectrum $\{0\}$.
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