= Solution
A path that reaches level $c$ from below does so while in state $+1$, because $Z$ is continuous and has slope $X$. At the first crossing time, the residual exponential holding time is memoryless and the <Strong Markov property> restarts the chain from $+1$. Conditional on ever reaching $c$, the probability of subsequently exceeding $c+d$ is therefore exactly the original probability of exceeding $d$. Hence
$$
\psi_+(c+d)=\psi_+(c)\psi_+(d),\qquad\psi_+(0+)=1.
$$
The zero-level limit is one because the initial $+1$ holding time is positive almost surely. Also $0<\psi_+(c)\leq1$, and it is nonincreasing. Consequently $h(c)=-\log\psi_+(c)$ is nondecreasing and additive. First $h$ is linear on positive rationals, then monotonicity gives the same result on all positive reals. Thus
$$
\boxed{\psi_+(c)=e^{-Ac}\text{ for some }A\geq0.}
$$
Absorption can be viewed as killing at rate $\lambda$, independent of sign changes at rate $q$; this ensures the maximum is finite almost surely and makes the positive decay rate found in (iv) natural.
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