= Solution
A <C-star algebra> is a complex <Banach algebra> with a conjugate-linear involution satisfying $(ab)^*=b^*a^*$ and $\|a^*a\|=\|a\|^2$. The inequality $\|a\|^2\le\|a^*\|\|a\|$, and then the same inequality for $a^*$, show that the involution is <isometric>. In a nonzero <unital> <C-star algebra>, the identity has <norm> one.
For a <Hermitian algebra element> $h$, the exponential $e^{ith}$ is unitary for real $t$ and has <norm> one. If $\lambda\in\sigma(h)$, then $e^{it\lambda}\in\sigma(e^{ith})$: factoring $e^{itz}-e^{it\lambda}=(z-\lambda)q(z)$ with entire $q$ proves this, since the evaluated factors commute. Hence $|e^{it\lambda}|\le1$ for all positive and negative $t$, forcing $\operatorname{Im}\lambda=0$. Thus \b[<Hermitian algebra elements> have real <Banach algebra spectrum>].
For a <normal C-star element> $a$, commutation of $a,a^*$ gives
$$
\|a^2\|^2=\|(a^*)^2a^2\|=\|(a^*a)^2\|=\|a^*a\|^2=\|a\|^4.
$$
The middle equality of <norms> uses the <C-star identity> on the <Hermitian algebra element> $a^*a$. Iterating along powers $2^n$ and applying the <spectral radius formula> proves $r(a)=\|a\|$.
Now suppose $A$ is commutative and <unital>. Every <maximal ideal> is <closed>: if it were dense, it would contain an element within distance less than one of the identity, hence an invertible element by the <Neumann series>, which is impossible for a proper ideal. Its quotient is a complex Banach division algebra and is therefore $\mathbb C$ by the <Gelfand-Mazur theorem>. <Maximal ideals> thus give <characters>. Conversely, every <character> $\varphi$ satisfies $\varphi(a)\in\sigma_A(a)$, because applying it to an inverse of $a-\varphi(a)1$ would give a contradiction. <Characters> are consequently contractive. The converse spectral inclusion follows by placing the proper ideal generated by $a-\lambda1$ in a <maximal ideal>. We have proved
$$
\sigma_A(a)=\{\varphi(a):\varphi\in\Phi_A\}.
$$
The <character space> $\Phi_A$ is a weak-star <closed> subset of the dual unit ball, characterized by $\varphi(1)=1$ and $\varphi(ab)=\varphi(a)\varphi(b)$. The <Banach-Alaoglu theorem> makes it <compact> Hausdorff. Define the <Gelfand transform> by $\widehat a(\varphi)=\varphi(a)$. For Hermitian $h$, its values are real; writing $a=h+ik$ with both terms Hermitian shows $\widehat{a^*}=\overline{\widehat a}$. In the commutative algebra every element is normal, so
$$
\|\widehat a\|_\infty=r(a)=\|a\|.
$$
The transform is therefore an <isometric> <unital> star-homomorphism and has <closed> range in $C(\Phi_A)$. Its range separates <characters> by their definition and is <closed> under complex conjugation. The complex <Stone-Weierstrass theorem> makes that range dense, hence equal to $C(\Phi_A)$. This proves the <Commutative Gelfand--Naimark theorem>:
$$
\boxed{A\cong C(\Phi_A)\quad\text{isometrically as a star algebra}.}
$$
For a nonunital <C-star algebra>, its <C-star unitization> can be constructed without assuming an operator representation on a <Hilbert space>. Represent $A$ by left multiplication on itself, which is <isometric> because testing $L_a$ on $a^*/\|a\|$ gives $\|L_a\|=\|a\|$. Adjoin the identity operator and <set> $\|z\|=\|L_z\|$. For $z=a+\lambda1$ and $b\in A$, the <C-star identity> in $A$ gives $\|zb\|^2=\|b^*z^*zb\|\le\|b\|\|L_{z^*z}b\|$. Consequently $\|L_z\|^2\le\|L_{z^*z}\|\le\|L_{z^*}\|\|L_z\|$. Applying this also to $z^*$ proves that the new involution is <isometric> and that its <C-star identity> holds. The extension by the identity is complete, since the <isometric> image of $A$ is <closed> and one adjoins just one dimension. Apply the <unital> result to a commutative $A^+$; the <character> $\infty$ given by $a+\lambda1\mapsto\lambda$ has kernel $A$. Under the isomorphism, this kernel consists of the <continuous> functions vanishing at $\infty$. Thus \b[$A\cong C_0(\Phi_A)$] in the nonunital case, with locally <compact> <character space> $\Phi_A=\Phi_{A^+}\setminus\{\infty\}$.
For a Hermitian $h$ in a possibly noncommutative algebra, apply this result to the commutative <unital> <C-star subalgebra> $C^*(1,h)$ in the <unitization>. Its <Banach algebra spectrum> at $h$ equals the ambient <Banach algebra spectrum>: both <Banach algebra spectra> are real, the complement of the ambient <compact> real <Banach algebra spectrum> is <connected>, and the resolvent-component result of Question 1 retains that whole complement in the smaller algebra. <Characters> of this <subalgebra> are determined by their values at $h$, so its <character space> is homeomorphic to $\sigma(h)$. We obtain the <continuous functional calculus> $C(\sigma(h))\to C^*(1,h)$.
Interpret $\mathbb R^+$ here as $[0,\infty)$. If $\sigma(h)\subseteq[0,\infty)$, apply this calculus to $t\mapsto\sqrt t$, obtaining a Hermitian $k$ with $k^2=h$. In a nonunital algebra this element remains in $A$ because the defining scalar function vanishes at zero. Conversely, if $h=k^2$ with $k=k^*$, <spectral mapping theorem> and the real <Banach algebra spectrum> of $k$ give $\sigma(h)=\{t^2:t\in\sigma(k)\}\subseteq[0,\infty)$. Hence \b[nonnegative <Banach algebra spectrum> is equivalent to being the square of a <Hermitian algebra element>]. A strict-positive interpretation would exclude $h=0$, and would not be equivalent to the assertion about an arbitrary Hermitian square.
Finally, use the two <continuous> real functions $t\mapsto\sqrt{\max(t,0)}$ and $t\mapsto\sqrt{\max(-t,0)}$ on $\sigma(x)$. Their calculus values $u,v$ are Hermitian, belong to $A$ also in the nonunital case, and satisfy
$$
\boxed{x=u^2-v^2.}
$$
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