= Solution
The time-changed diffusion has generator $\mathcal A=(\kappa/2)\partial_{zz}+2(1/z+1/(z-1))\partial_z$. For $1<z<R$, its <boundary hitting probability from a diffusion scale function> is
$$
\mathbb P_z(\tau_R<\tau_1)=\frac{s_\kappa(z)}{s_\kappa(R)}.
$$
To see this directly, stop the scale <local martingale> on $[1+\epsilon,R]$ and pass to $\epsilon\downarrow0$. Exit from $(1,R)$ occurs in finite clock time: the scale density behaves as $(z-1)^{-4/\kappa}$ and the speed density as $(z-1)^{4/\kappa}$, so the usual scale-times-speed exit integral is finite at $1$. More explicitly, with $m(v)=2/(\kappa s_\kappa'(v))$ the <finite-interval diffusion exit Green kernel> gives
$$
\mathbb E_z\tau_{1,R}=\int_1^R
\frac{s_\kappa(z\wedge v)[s_\kappa(R)-s_\kappa(z\vee v)]}{s_\kappa(R)}\,m(v)\,dv.
$$
The equation $\mathcal A u=-1$ follows by differentiation and the derivative jump at $v=z$, so stopping the corresponding Itô identity gives this exit bound. The integrand near $1$ is $O(v-1)$; elsewhere it is bounded. At large $z$ the drift is bounded, excluding explosion at infinity in finite clock time.
Here is why these exits describe the original swallowing event. In clock time,
$$
D(u)=D(0)\exp\left(-2\int_0^u\frac{dv}{\widehat Z_v(\widehat Z_v-1)}\right).
$$
If $\widehat Z$ hits $1$ at finite $u$, its integrated <stochastic differential equation> shows $\int_0^{\tau_1}(\widehat Z_v-1)^{-1}dv<\infty$: the Brownian term has a finite limit and both positive drift integrals must then have finite limits. Consequently $D(\tau_1)>0$, $X$ hits zero and $Y$ stays positive. This gives $T(x)<T(y)$.
Conversely $T(x)$ is finite by part (b). If its clock ends at a finite value, the diffusion must hit $1$; an interior endpoint with positive $D$ would leave $X$ positive. If the clock runs forever without hitting $1$, the standard one-dimensional scale classification gives escape to infinity whenever $S_\kappa<\infty$. This cannot represent a strict swallowing event, which has $D\to Y_{T(x)}>0$ and $Z\to1$. Hence the surviving event represents $T(x)=T(y)$. This reasoning also explains why a simultaneous collision must be treated separately rather than assigned the boundary value at $1$.
For $\kappa\geq8$, the scale is unbounded. The <maximal inequality for a nonnegative supermartingale> gives $\mathbb P(\sup s_\kappa(\widehat Z)\geq s_\kappa(R))\leq s_\kappa(z)/s_\kappa(R)$; letting $R\to\infty$ rules out escape. Finite-interval exit then forces a hit at $1$. For $4<\kappa<8$, taking $R\to\infty$ gives
$$
\boxed{F(x,y)=1-\frac{s_\kappa(y/(y-x))}{S_\kappa},\qquad
\mathbb P(T(x)=T(y))=\frac{s_\kappa(y/(y-x))}{S_\kappa}>0.}
$$
The strict-event probability is also positive, since $0<s_\kappa(z)<S_\kappa$ for finite $z$.
For $\kappa\geq8$, first intersect the probability-one strict-order events over rational $0<a<b$. The real boundary flow gives nondecreasing swallowing times on $(0,\infty)$. For any $0<x<y$, choose rational $x<a<b<y$; then $T(x)\leq T(a)<T(b)\leq T(y)$. This proves the simultaneous assertion
$$
\boxed{\text{almost surely }T(x)<T(y)\text{ for every }0<x<y,\quad\kappa\geq8.}
$$
The range is positive points, as in the definition of $F$; reflection reverses the ordering on the negative axis. At the printed endpoint $\kappa=4$, each fixed positive point has $T=\infty$. Thus $\mathbb P(T(x)=T(y))=1$ if equality of extended times is allowed, but there is no finite simultaneous-swallowing assertion there. The preceding formulas assume $\kappa>4$.
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