= Solution
Let $K=\sup_x f(x)<\infty$ and let $C$ be a <Lipschitz constant> for $f$. Since
$$
|f(x)^2-f(y)^2|\leq2KC|x-y|,
$$
both coefficients $f$ and $-f^2/2$ are bounded and <globally Lipschitz>. The <global existence theorem for stochastic differential equations with Lipschitz coefficients> gives a unique global <strong solution of a stochastic differential equation>.
The same <Itô formula> cancellation as before gives
$$
d(e^{X_t})=e^{X_t}f(X_t)\,dB_t,\qquad
[e^X]_t=\int_0^te^{2X_s}f(X_s)^2\,ds.
$$
For $S=T_{-a}\wedge T_b$, the stopped exponential is bounded between $e^{-a}$ and $e^b$. Continuity and strict positivity of $f$ give
$$
c_{a,b}:=\min_{x\in[-a,b]}e^{2x}f(x)^2>0.
$$
The <Itô isometry> therefore yields
$$
1+c_{a,b}\mathbb E[t\wedge S]
\leq\mathbb E e^{2X_{t\wedge S}}\leq e^{2b}.
$$
The <monotone convergence theorem> proves $\mathbb ES<\infty$. The bounded <optional stopping> argument is now identical to part (b)(iii), and gives
$$
\mathbb P(T_b<T_{-a})=\frac{1-e^{-a}}{e^b-e^{-a}},\qquad
\mathbb P(T_b<\infty)=e^{-b}.
$$
Passing from level hitting to strict exceedances as in part (b)(iv),
$$
\boxed{X^\star\sim\operatorname{Exp}(1).}
$$
No global positive lower bound on $f$ was needed: compact-interval positivity suffices for each exit argument. The law is independent of the particular $f$ because the <scale function> is always $s(x)=e^x$: its <infinitesimal generator> satisfies
$$
\mathcal Ls=\frac{f^2}{2}(s''-s')=0.
$$
This is the <exponential scale function for a drift tied to diffusivity>. Changing $f$ changes how quickly the path is traversed, but leaves these boundary probabilities and the maximum law unchanged.
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