= Solution
Write $X_t=x+W_t+\mu t$ with $W$ a zero-starting <Brownian motion>. A unit-time increment exceeds $2b$ with probability
$$
p=\mathbb P(N(\mu,1)>2b)>0.
$$
If the process has not exited before an integer time $k$, it is in $(-b,b)$; an increment exceeding $2b$ forces it outside at time $k+1$ and hence forces an exit by then. Independence of <Brownian increments> gives $\mathbb P(T>k)\le(1-p)^k$. Thus the <geometric tail bound from a uniform escape probability> yields
$$
\mathbb E T^2=\int_0^\infty2s\,\mathbb P(T>s)ds
\le\sum_{k=0}^\infty(2k+1)(1-p)^k=\frac{2-p}{p^2}<\infty.
$$
In particular $T$ is finite almost surely.
The <diffusion generator> is $\mathcal L=\tfrac12\partial_{xx}+\mu\partial_x$. The function $f(z)=e^{-2\mu z}$ satisfies $\mathcal Lf=0$, so the <Itô formula> gives $df(X_t)=-2\mu f(X_t)dW_t$. It is a true <martingale>, since it is $e^{-2\mu x}$ times the Gaussian exponential <martingale> $e^{-2\mu W_t-2\mu^2t}$. The stopped values $f(X_{t\wedge T})$ are uniformly bounded because the stopped path stays in $[-b,b]$. Optional sampling and bounded convergence give
$$
e^{-2\mu x}=p_-e^{2\mu b}+(1-p_-)e^{-2\mu b},
$$
where continuity ensures that $X_T$ is exactly one of the two endpoints. Solving,
$$
\boxed{\mathbb P(T_-<T_+)=\frac{e^{-2\mu x}-e^{-2\mu b}}{e^{2\mu b}-e^{-2\mu b}}.}
$$
This <drifted Brownian interval-exit probability> tends to $(b-x)/(2b)$ as $\mu\to0$, as expected.
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