= Solution
Put $D_t=B_t-B'_t$. By <independence> of the two <Brownian motions>, $D_t/\sqrt2$ is a standard <Brownian motion>: its <independent increments> have the <normal distribution> with <variance> equal to elapsed time. The <stopping time> $\tau$ is the first time $D$ hits $-1$ or $3$.
To justify finiteness without presupposing the desired <hitting probability>, each unit-time increment of $D$ has the <normal distribution> $N(0,2)$. An increment of absolute value greater than $4$ forces exit from $[-1,3]$ if the path was inside at its start. These <independent increments> give $\mathbb P(\tau>n)\leq(1-p)^n$, where $p=\mathbb P(|N(0,2)|>4)>0$. Hence $\tau<\infty$ <almost surely>.
The <stopped martingale> $D_{t\wedge\tau}$ lies in $[-1,3]$. The <optional stopping theorem> at $t\wedge\tau$ gives $\mathbb ED_{t\wedge\tau}=0$, and the <bounded convergence theorem> gives $\mathbb ED_\tau=0$. Writing $q=\mathbb P(D_\tau=-1)$, path <continuity> gives $0=-q+3(1-q)$. Therefore, as in <Brownian exit from an interval>,
$$
\boxed{\mathbb P(B_\tau=B'_\tau-1)=\frac34.}
$$
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