= Solution
Write $K=\sigma_A(x)$ and equip $\mathcal O(U)$ with the <compact-open topology>. All <algebra homomorphisms> here are complex linear. Choose a finite positively oriented <boundary> cycle $\Gamma$ in $U\setminus K$ with <winding number> one on a neighborhood of $K$ and zero outside $U$. Such a cycle can be obtained as the <boundary> of a finite union of sufficiently small squares whose <interior> contains $K$ and whose closure is contained in $U$; holes are included with negative <boundary> orientation. Define the <holomorphic functional calculus> by the norm-convergent <contour integral>
$$
\Theta_x(f)=\frac1{2\pi i}\int_\Gamma f(z)R(z)\,dz,
\qquad R(z)=(z1-x)^{-1}.
$$
The <resolvent identity> proves that $R$ is a <Banach-space-valued holomorphic function> off $K$. The <Cauchy integral theorem>, applied after each <bounded linear functional>, makes the integral independent of the chosen <boundary> cycle. The <Hahn-Banach theorem> then gives equality of the algebra-valued integrals. Moreover,
$$
\|\Theta_x(f)\|\le\frac{\operatorname{length}(\Gamma)}{2\pi}\sup_\Gamma\|R(z)\|\sup_\Gamma|f(z)|,
$$
which proves complex <linearity> and <continuity> for the <compact-open topology>.
For the constant function $1$ and the coordinate function $Z$, the contour can be deformed to a large circle because their integrands are <holomorphic> everywhere outside $K$. There the <Neumann series> gives $R(z)=\sum_{n\ge0}x^nz^{-n-1}$. Integrating term by term shows
$$
\Theta_x(1)=1,\qquad\Theta_x(Z)=x.
$$
To prove multiplicativity, take nested cycles $\Gamma_o,\Gamma_i$ around $K$, with the outer cycle winding once on the inner cycle and the inner winding zero on the outer cycle. Use the outer cycle for $f$ and the inner for $g$. The <resolvent identity> reads
$$
R(z)R(w)=\frac{R(w)-R(z)}{z-w}.
$$
In the resulting double integral, the term containing $R(w)$ gives $f(w)g(w)R(w)$ after the $z$ integral, by the <Cauchy integral formula>. The term containing $R(z)$ vanishes after the $w$ integral, because $w\mapsto g(w)/(z-w)$ is <holomorphic> on and inside the inner cycle's region. Therefore
$$
\Theta_x(f)\Theta_x(g)=\Theta_x(fg).
$$
The integrals are over finite <compact> curves and have <continuous> integrands, so exchanging their order is legitimate. This proves the existence of the required <continuous> unital <algebra homomorphism>.
For uniqueness, the <Runge approximation theorem> gives density in $\mathcal O(U)$, for the <compact-open topology>, of <rational functions> with poles outside $U$. More explicitly, exhaust $U$ by <compact> sets that are holomorphically convex relative to $U$, and approximate on those sets by <rational functions> whose poles lie in the complement of $U$; this gives convergence on every <compact> subset. Any unital <algebra homomorphism> taking $Z$ to $x$ must take $(Z-\alpha)^{-1}$, for $\alpha\notin U$, to $(x-\alpha1)^{-1}$, since it preserves inverse equations. Partial fractions and <polynomial> algebra therefore fix its value on every such <rational function>. <Continuity> fixes its value on their limits. Thus \b[the <continuous> unital homomorphism with the prescribed coordinate value is unique]. Restricting a function to a smaller neighborhood of $K$ gives the same value, either by the contour definition or this uniqueness argument.
To prove the <holomorphic spectral mapping theorem>, first let $\mu\notin f(K)$. There is an open neighborhood $V\subseteq U$ of $K$ on which $f-\mu$ has no zeros. The <holomorphic function> $h=1/(f-\mu)$ on $V$ supplies a two-sided inverse through
$$
\Theta_x(f)-\mu1=\Theta_x(f-\mu),\qquad
\Theta_x(f-\mu)\Theta_x(h)=\Theta_x(h)\Theta_x(f-\mu)=1.
$$
Thus $\sigma_A(\Theta_x(f))\subseteq f(K)$. Conversely, if $\lambda\in K$ and $\mu=f(\lambda)$, the divided difference
$$
g(z)=\frac{f(z)-f(\lambda)}{z-\lambda},\qquad g(\lambda)=f'(\lambda),
$$
is <holomorphic> on $U$, and multiplicativity gives
$$
\Theta_x(f)-\mu1=(x-\lambda1)\Theta_x(g).
$$
The two factors commute. A commuting product $uv$ can be invertible only if both factors are invertible: its inverse commutes with each factor, and $v(uv)^{-1}$ is a two-sided inverse for $u$. Invertibility of the displayed product would therefore contradict $\lambda\in K$. This proves
$$
\boxed{\sigma_A(\Theta_x(f))=f(\sigma_A(x)).}
$$
For the root, let $D=\mathbb C\setminus(-\infty,0]$ and $q(z)=\exp(\operatorname{Log}z/3)$, using the <principal complex logarithm>. Define $y=\Theta_x(q)$. Since $q(z)^3=z$, multiplicativity and the <holomorphic spectral mapping theorem> give
$$
\boxed{y^3=x,\qquad\sigma_A(y)=q(K)\subseteq S,\qquad S=\{z\ne0:|\arg z|<\pi/3\}.}
$$
This is the <principal cube root of a Banach-algebra element>.
For uniqueness, suppose $v^3=x$ and $\sigma_A(v)\subset S$. Cubing maps the whole open sector $S$ into $D$, so
$$
F:\mathcal O(D)\longrightarrow A,\qquad F(h)=\Theta_v\bigl(z\mapsto h(z^3)\bigr)
$$
is a <continuous> unital <algebra homomorphism>: pullback by cubing is <continuous> for the <compact-open topology>, since images of <compact> sets are <compact>. It takes the coordinate function to $v^3=x$, and uniqueness already proved gives $F=\Theta_x$. For $z\in S$, the scalar <principal cube root> satisfies $q(z^3)=z$, because $3\arg z\in(-\pi,\pi)$. Hence
$$
v=\Theta_v(q\circ(z\mapsto z^3))=F(q)=\Theta_x(q)=y.
$$
This proves \b[uniqueness with the stated spectral sector], without assuming the ambient <Banach algebra> is <commutative> or that <algebra characters> separate its elements.
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