= Solution
Assume the finite-generator identity holds whenever there is no common zero. Let $\varphi\in\Phi_A$. For every finite family $f_1,\ldots,f_n\in\ker\varphi$, their zero sets must have nonempty intersection: otherwise an identity $\sum f_jg_j=1$ would give $0=\varphi(1)=1$. These zero sets are closed in compact $K$. The <finite intersection property> therefore supplies $x\in K$ at which every member of $\ker\varphi$ vanishes. For any $f\in A$, the element $f-\varphi(f)1$ is in that kernel, so $f(x)=\varphi(f)$. Thus $\varphi=\delta_x$, proving
$$
\boxed{\text{the finite-generator identity holds}\iff A\text{ is natural}.}
$$
For $C(K)$, the identity is explicit. If the $f_j$ have no common zero, the continuous function $q=\sum_j|f_j|^2$ is strictly positive on compact $K$. Set
$$
g_j=\frac{\overline{f_j}}{q}\in C(K).
$$
Then $\sum f_jg_j=1$. The equivalence just proved shows that \b[$C(K)$ is natural].
Next consider the <disc algebra> $A(\overline{\mathbb D})$, the continuous functions on the closed unit disk that are holomorphic in its interior. We show directly that each <algebra character> is evaluation. Polynomials are uniformly dense in this algebra: for $0<r<1$, $f_r(z)=f(rz)$ tends uniformly to $f$ as $r\uparrow1$, by <uniform continuity>. The function $f_r$ is holomorphic on $|z|<1/r$, so its Taylor polynomials converge uniformly on $|z|\leq1$.
Every <algebra character> of a unital <Banach algebra> is bounded, with $|\varphi(f)|\leq\|f\|$: if $f-\varphi(f)1$ were invertible, applying $\varphi$ would contradict invertibility, so $\varphi(f)$ belongs to the <spectrum of an element>, which is contained in the disk of radius $\|f\|$ by the <Neumann series>. For the coordinate function $z$, put $\lambda=\varphi(z)$; then $|\lambda|\leq1$. Multiplicativity gives $\varphi(p)=p(\lambda)$ for every polynomial. By polynomial density and continuity, $\varphi(f)=f(\lambda)$ for every member of the <disc algebra>. Hence
$$
\boxed{\Phi_{A(\overline{\mathbb D})}\cong\overline{\mathbb D},\qquad A(\overline{\mathbb D})\text{ is natural}.}
$$
Finally consider the algebra $C$ in the last request. For each $f\in C$, its holomorphic extension $g$ from the boundary is unique by the <maximum modulus principle>. Define $T(f)=g\in A(\overline{\mathbb D})$. This is a surjective unital algebra homomorphism, since it restricts to the identity on the <disc algebra>. Also
$$
\|T(f)\|_\infty=\sup_{|z|=1}|f(z)|\leq\|f\|_\infty.
$$
The algebra $C$ is closed: if $f_n\to f$ uniformly, the associated $T(f_n)$ are uniformly Cauchy by this bound, so their limit belongs to the closed <disc algebra> and agrees with $f$ on the boundary. Constants and the coordinate function belong to $C$, so it is itself a <uniform algebra> on the disk. Its <ideal>
$$
I=\ker T=\{f\in C(\overline{\mathbb D}):f|_{\partial\mathbb D}=0\}
$$
is contained in $C$, and $C/I$ is the <disc algebra>.
If a <algebra character> $\chi$ on $C$ vanishes on $I$, it factors through $T$ and the preceding result gives
$$
\chi(f)=T(f)(\lambda)\quad\hbox{for some }\lambda\in\overline{\mathbb D}.
$$
If $\chi$ does not vanish on $I$, choose $h\in I$ with $\chi(h)\ne0$. For $F\in C(\overline{\mathbb D})$, define
$$
\widetilde\chi(F)=\frac{\chi(hF)}{\chi(h)}.
$$
The product $hF$ is in $I$, so this is well defined. It is unital and linear, and
$$
\chi(hF)\chi(hG)=\chi(h^2FG)=\chi(h)\chi(hFG)
$$
shows multiplicativity. Since $C(\overline{\mathbb D})$ is natural, $\widetilde\chi$ is evaluation at a point $x$ of the closed disk. The equality $h(x)=\widetilde\chi(h)=\chi(h)\ne0$ places $x$ in the open disk. On $C$, $\widetilde\chi(f)=\chi(f)$, so $\chi$ itself is evaluation at that interior point.
We therefore obtain two families: actual evaluations $f\mapsto f(x)$ on one closed disk, and analytic-extension evaluations $f\mapsto T(f)(\lambda)$ on a second closed disk. They agree on the boundary. They are distinct at interior points: functions in $I$ distinguish actual interior evaluations from every analytic-extension evaluation, and the coordinate function distinguishes points within each family. Thus
$$
\boxed{\Phi_C\cong\overline{\mathbb D}\ \cup_{\partial\mathbb D}\ \overline{\mathbb D}\cong S^2.}
$$
This is also a topological identification. Each disk family is continuous in the <Gelfand topology>, and they agree along their boundary, giving a continuous <bijection> from the glued disks to $\Phi_C$. The glued disks form a compact space, explicitly homeomorphic to the sphere by sending $(x,y)$ in the two copies to $(x,y,\pm\sqrt{1-x^2-y^2})$. Since the <character space> is Hausdorff, the continuous <bijection> is a <homeomorphism>. This <boundary-analytic disk algebra with doubled character space> has more <algebra characters> than evaluations on its original disk.
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