= Solution
\b[A bounded-domain hypothesis is needed for the requested global construction and uniqueness.] The printed question does not include it. For example, the upper half-plane satisfies the stated <exterior cone condition>, but both $u=0$ and $u(x_1,x_2)=x_2$ are <harmonic functions> with zero <Dirichlet boundary condition>. Thus part (iv), in its printed unrestricted solution class, is false. In this question's solutions we add that $\Omega$ is bounded; the <exterior cone condition> and all other data remain as printed.
The half-plane also rules out the printed global separated barrier. In its angular interval $(0,\pi)$, varying $r$ on a fixed ray forces $\nu=2$ if $r^{\nu-2}(h''+\nu^2h)\geq\delta>0$ for every $r>0$. We would then have $h''+4h\geq\delta$ and $h<0$. Put $a=\pi/4$ and $b=3\pi/4$. Twice <integration by parts> gives
$$
\int_a^b(h''+4h)\sin(2(\theta-a))\,d\theta=2\bigl(h(a)+h(b)\bigr)<0,
$$
contradicting the positive integrand. Thus a source qualification is necessary for part (i) as well as part (iv).
Fix $x_0$ and its exterior cone of half-angle $\alpha$. Choose $0<\alpha_0<\min(\alpha,\pi/2)$ and set
$$
\beta=\frac{\pi}{2(\pi-\alpha_0)},\qquad 0<\nu<\beta<1.
$$
On the complement of the cone, unwrap the angle in <polar coordinates> as $\vartheta\in(\alpha,2\pi-\alpha)$, measured from the cone axis. Then $\psi=\vartheta-\pi$ is the angle from the opposite axis and $|\psi|\leq\pi-\alpha$ on $\overline\Omega\setminus\{x_0\}$. Define the <power barrier for an exterior cone> by
$$
\boxed{g(r,\vartheta)=-r^\nu\cos\bigl(\beta(\vartheta-\pi)\bigr),\qquad g(x_0)=0.}
$$
This has the requested separated form $r^\nu h(\vartheta)$. In the printed signed-angle convention, the same function is $-r^\nu\cos(\beta(\pi-|\theta|))$ outside the cone. It is smooth across the negative axis: near that axis the angular expression is the even function $\cos(\beta\psi)$.
Since $\beta(\pi-\alpha)<\pi/2$, put $m=\cos(\beta(\pi-\alpha))>0$. The cosine is at least $m$, so $g\leq-mr^\nu<0$ away from $x_0$, and $g$ is <continuous> at $x_0$. The <Laplacian in polar coordinates> gives
$$
\Delta g=(\beta^2-\nu^2)r^{\nu-2}\cos(\beta\psi).
$$
Let $D=\sup_{x\in\Omega}|x-x_0|<\infty$. Since $\nu<2$, the <barrier for the Dirichlet problem> satisfies
$$
\boxed{\Delta g\geq\delta:=(\beta^2-\nu^2)mD^{\nu-2}>0.}
$$
Both the uniform lower bound and the later global comparison use boundedness. In particular, $-g$ is positive at every other boundary point, and is bounded away from zero on boundary sets staying a positive distance from $x_0$.
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