= Solution
Write $K=\sigma_A(x)$ and $R_x(\zeta)=(\zeta1-x)^{-1}$. Suppose first that a unital complex-algebra homomorphism sends $Z$ to $x$. If $\lambda\notin U$, the <holomorphic function> $1/(Z-\lambda)$ belongs to $\mathcal O(U)$ and is the multiplicative inverse of $Z-\lambda$. Its image is an inverse for $x-\lambda1$. Thus \b[existence forces $K\subset U$], even before continuity is used.
For the converse, assume $K\subset U$. Choose a bounded finite union of polygonal regions $V$ with $K\subset V$, $\overline V\subset U$, and boundary avoiding $K$. Orient its boundary cycle $\Gamma$ positively around $V$, including negative orientations on holes. Its winding number is one near $K$ and zero outside $U$. Define
$$
\boxed{\Theta_x(f)=\frac1{2\pi i}\int_\Gamma f(\zeta)R_x(\zeta)\,d\zeta.}
$$
The integral exists in $A$ because the integrand is continuous on finitely many compact contour pieces. The <Cauchy theorem> shows it is unchanged by replacing $\Gamma$ with another such spectral cycle: the difference has zero winding around the possible singularities in $K$. Banach-valued Cauchy statements follow by applying <bounded linear functionals>. We use the scalar <Cauchy integral formula> in its cycle form: a cycle with index one at a point and surrounding only a holomorphy region integrates $f(\zeta)/(\zeta-z)$ to $2\pi i f(z)$.
Linearity is immediate and, for a fixed cycle,
$$
\|\Theta_x(f)\|\le\frac{\operatorname{length}\Gamma}{2\pi}
\max_\Gamma\|R_x(\zeta)\|\,\sup_\Gamma|f|.
$$
This is a bound by one compact-uniform seminorm, proving continuity for <local uniform convergence>. To obtain $\Theta_x(1)=1$, deform the resolvent integral, which is holomorphic on $\mathbb C\setminus K$, to a large circle and integrate its <Neumann series>. Similarly $\zeta R_x(\zeta)=1+xR_x(\zeta)$ gives $\Theta_x(Z)=x$.
For multiplicativity choose an outer spectral cycle for $f$ and an inner one for $g$, nested in $U$ so the inner cycle lies inside the outer region. The <resolvent identity> is
$$
R_x(\zeta)R_x(\eta)=\frac{R_x(\eta)-R_x(\zeta)}{\zeta-\eta}.
$$
Substitute it into the double integral for $\Theta_x(f)\Theta_x(g)$. In the term containing $R_x(\eta)$, first integrate $f(\zeta)/(\zeta-\eta)$ around the outer cycle, obtaining $f(\eta)$. In the term containing $R_x(\zeta)$, first integrate $g(\eta)/(\zeta-\eta)$ around the inner cycle, obtaining zero because $\zeta$ lies outside it. Interchange is valid for the continuous integrands on disjoint compact cycles. Hence
$$
\Theta_x(f)\Theta_x(g)=\Theta_x(fg).
$$
This proves the required continuous unital homomorphism, the <holomorphic functional calculus>.
For uniqueness use the following version of the allowed <Runge theorem>: on any open $U\subseteq\mathbb C$, <rational functions> with all finite poles outside $U$ approximate each <holomorphic function> locally uniformly; poles at infinity are allowed, giving <polynomials>. A unital homomorphism sending $Z$ to $x$ is forced on <polynomials> and on every inverse $(Z-\lambda)^{-1}$, $\lambda\notin U$, and hence on these <rational functions>. Continuity then forces its value on their limits. Thus \b[$\Theta_x$ is unique]. This argument works for disconnected $U$ as well and does not assume <polynomials> alone are dense. It proves <continuity and uniqueness of holomorphic functional calculus>.
We will use the character description of <spectra>, with a short justification. A <character of an algebra> $\varphi$ is a nonzero complex multiplicative <linear functional>; it satisfies $\varphi(1)=1$. Applying it to an inverse shows $\varphi(a)\in\sigma_A(a)$, so $|\varphi(a)|\le\|a\|$ and the character is automatically continuous. Conversely, for noninvertible $a-\lambda1$, its <principal ideal> is proper because $A$ is commutative. Put it in a <maximal ideal> $M$. The closure of $M$ is proper: an ideal containing an element sufficiently close to one contains an invertible element and thus the identity. Maximality makes $M$ closed. The quotient is a complex Banach division algebra. By the nonempty-spectrum result of Question 3, each quotient element differs from some scalar by a noninvertible element, which in a division algebra must be zero. Thus the quotient is $\mathbb C$ and its quotient map is a character taking $a$ to $\lambda$. This proves
$$
\sigma_A(a)=\{\varphi(a):\varphi\text{ a character of }A\}.
$$
It is the <spectrum equals character values in a commutative Banach algebra> statement and does not assume semisimplicity.
For any character, continuity permits passing it inside the contour integral. Since $\varphi(R_x(\zeta))=(\zeta-\varphi(x))^{-1}$ and $\varphi(x)\in K$, the scalar Cauchy formula gives
$$
\boxed{\varphi(\Theta_x(f))=\frac1{2\pi i}\int_\Gamma
\frac{f(\zeta)}{\zeta-\varphi(x)}\,d\zeta=f(\varphi(x)).}
$$
Apply the character description first to $\Theta_x(f)$ and then to $x$ to obtain the <holomorphic spectral mapping theorem>
$$
\boxed{\sigma_A(\Theta_x(f))=f(\sigma_A(x)).}
$$
On $\Pi_+$, use the analytic logarithm with argument in $(-\pi/2,\pi/2)$ and set $s(z)=\exp(\tfrac12\operatorname{Log} z)$. It satisfies $s(z)^2=z$ and takes its values in $\Pi_+$. The <principal square root in a commutative Banach algebra> is
$$
\boxed{y=\Theta_x(s),\qquad y^2=x,\qquad\sigma_A(y)\subset\Pi_+.}
$$
For uniqueness let $v$ be another root with <spectrum> in $\Pi_+$. For each character, $\varphi(y)$ and $\varphi(v)$ are scalar roots of $\varphi(x)$ with positive real part, so both equal $s(\varphi(x))$. Hence every character takes the nonzero value $2s(\varphi(x))$ at $y+v$. The character description implies $y+v$ is invertible. Commutativity gives $(y-v)(y+v)=y^2-v^2=0$, and multiplying by the inverse yields $y=v$. Equal character values alone would not imply equality in a possibly nonsemisimple algebra; the invertible sum is the essential extra step.
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