Let $T_b = \inf \{t \ge 0: W_t = b\}$ for brownian motion $W$ and $P^x$ will be a one-dimensional Brownian family.
Then we have
Show that for $a>0, 0 \le x \le a:$ $$P^x[T_0 < T_a] = \frac{a-x}{a}, P^x[T_a < T_0] = \frac{x}{a}.$$
So to get the two probabilities, we just need to integrate the two infinite sums over $t$. However, I don't know how to evaluate the integrals for these sums.
Also, how can we use this to compute $E^x(T_0 \wedge T_a) = x(a-x), 0\le x \le a$? I would greatly appreciate any help.
My idea would be the following (I have not checked it through, but you could try it at your own at first):
By dominated convergence we have $$\Bbb P^x (T_0 < T_a) = \lim_{\alpha \to 0 } \Bbb E[ e^{-\alpha T_0 \wedge T_a} 1_{T_0 < T_a}] = \lim_{\alpha \to 0} \frac{\sinh ((a-x)\sqrt{2\alpha})}{\sinh (a\sqrt{2\alpha})}$$ This limit should be computable (looks like l'Hospital). Note that $\Bbb P^x (T_a < T_0) = 1 - \Bbb P^x (T_0 < T_a)$.
Further we should have
$$\Bbb E [T_0 \wedge T_a ]= \lim_{\alpha \to 0} \Bbb E [T_0 \wedge T_a e^{-\alpha T_0 \wedge T_a}] = \lim_{\alpha \to 0} - \frac{\text d }{\text d \alpha} \Bbb E[ e^{-\alpha T_0 \wedge T_a}] = \lim_{\alpha \to 0} - \frac{\text d }{\text d \alpha} \frac{(x -\frac a 2)\sqrt{2\alpha}}{\cosh(\frac a 2 \sqrt{2\alpha})}$$
You could share the exact computations as own answer, if this works and you have figured it out.
By the way, here is the martingale argument:
$M_t = W_{t\wedge T_0 \wedge T_a} - W_0$ is a martingale, thus by dominated convergence $$ x = \lim_{t\to \infty}\Bbb E^x [M_t] = \Bbb E^x [W_{T_0 \wedge T_a}] = a\Bbb P^x(T_a < T_0)$$