= Solution
For real $\lambda$, the <normal> moment-generating formula gives $\mathbb E e^{\lambda(B_t-B_s)}=e^{\lambda^2(t-s)/2}$. Using <independent increments> and conditioning on the Brownian <filtration>,
$$
\mathbb E\left[e^{\lambda B_t-\lambda^2t/2}\mid\mathcal F_s\right]
=e^{\lambda B_s-\lambda^2t/2}e^{\lambda^2(t-s)/2}
=e^{\lambda B_s-\lambda^2s/2}.
$$
All terms are integrable by the same <normal> formula. Thus the displayed process is an <Exponential martingale for Brownian motion>.
For the interval exit time, we first show both almost-sure finiteness and finite expectation. On $\{T>n\}$, the position at time $n$ lies in $(-b,a)$. If $X_{n+1}-X_n\ge a+b$, the continuous path must exit by time $n+1$. This increment is <independent> of $\mathcal F_n$ and has law $N(\mu,1)$, so the conditional probability of this event is a fixed $c>0$. Therefore
$$
\mathbb P(T>n+1)\le(1-c)\mathbb P(T>n),
\qquad \mathbb P(T>n)\le(1-c)^n.
$$
This <geometric tail bound from a uniform escape probability> implies $T<\infty$ almost surely and $\mathbb ET<\infty$.
Take $\lambda=-2\mu$. Then $e^{-2\mu X_t}=e^{-2\mu B_t-2\mu^2t}$ is the exponential <martingale> just proved. Its stopped values lie between $e^{-2\mu a}$ and $e^{2\mu b}$. The <optional stopping theorem> at $T\wedge t$, followed by bounded convergence, gives $\mathbb E e^{-2\mu X_T}=1$. With $v=\mathbb P(X_T=a)$ and the two possible boundary values,
$$
v e^{-2\mu a}+(1-v)e^{2\mu b}=1.
$$
Solving yields
$$
\boxed{\mathbb P(X_T=a)=\frac{1-e^{-2\mu b}}{1-e^{-2\mu(a+b)}}.}
$$
Next apply bounded-time <optional stopping> to $B_t$. Since $B_{T\wedge t}=X_{T\wedge t}-\mu(T\wedge t)$, we obtain $\mathbb EX_{T\wedge t}=\mu\mathbb E(T\wedge t)$. The stopped position is bounded, and the stopped time increases to an integrable $T$. Passing to the limit gives $\mu\mathbb ET=av-b(1-v)$. Therefore
$$
\boxed{\mathbb ET=\frac1\mu\left[(a+b)\frac{1-e^{-2\mu b}}{1-e^{-2\mu(a+b)}}-b\right].}
$$
As a check, letting $\mu\downarrow0$ gives the driftless formulas $v\to b/(a+b)$ and $\mathbb ET\to ab$. The expectation calculation uses a justified stopping limit, rather than assuming that <optional stopping> holds at every unbounded time.
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