= Solution
Given $0<z<1$, take
$$
w=\frac{1+\sqrt{1-z^2}}z>1,
\qquad z=\frac2{w+w^{-1}}.
$$
Choose $A,B$ so that
$$
Aw^{-a}+Bw^a=1,\qquad
Aw^b+Bw^{-b}=1.
$$
Solving,
$$
A=\frac{w^a-w^{-b}}{w^{a+b}-w^{-(a+b)}},
\qquad
B=\frac{w^b-w^{-a}}{w^{a+b}-w^{-(a+b)}}.
$$
The stopped discounted martingale is bounded, and $T$ is finite, so part (b) gives
$$
A+B=\mathbb E\!\left[
z^T(Aw^{X_T}+Bw^{-X_T})\right]=\mathbb E(z^T).
$$
Therefore the <discounted symmetric random-walk exit transform> is
$$
\boxed{
\mathbb E(z^T)=
\frac{w^a+w^b-w^{-a}-w^{-b}}
{w^{a+b}-w^{-(a+b)}}},
\qquad
w=\frac{1+\sqrt{1-z^2}}z.
$$
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