$(X_k)_{k \in \mathbb{N}}$ is iid with $\mathbf{P} [X_1 = \frac{1}{2}] = \mathbf{P} [X_1 = \frac{3}{2}] = \frac{1}{2}$. $M = (M_n)_{n \in \mathbb{N}_0}$ with $M_0 = 1$ and $M_n = \prod_{k = 1}^n X_k$ for $n > 0$ is a martingale.
For $0 < a < \frac{1}{2}$ and $b > 2$, $\tau := \inf_{n \in \mathbb{N}_0} \{ M_n \ge b \vee M_n \le a \} = \tau_a \wedge \tau_b$ is a stopping time with $\mathbf{P} [\tau < \infty] = 1$. Show that
$\frac{2 - 2 a}{3b - 2a} < \mathbf{P} [M_\tau \ge b] < \frac{2 - a}{2 b - a}$.
Thoughts: With the canonical symmetric random walk, I would use $\mathbf{P} [\tau < \infty] = 1$ (i.e., $M$ stops almost surely by hitting $a$ or $b$) to argue $\mathbf{E} [M_0] = \lim_{n \rightarrow \infty} \mathbf{E} [M_{\tau \wedge n}] = \mathbf{E} [M_\tau] = a \mathbf{P} [\tau = \tau_a] + b \mathbf{P} [\tau = \tau_b]$ and solve for $\mathbf{P} [\tau = \tau_b]$. In fact, it seems to me that $\mathbf{P} [\tau = \tau_b] = \mathbf{P} [M_\tau \ge b]$. However, $M$ can take all kinds of values (greater than $b$) at the time when it crosses $b$, so this doesn't work here.
You are right that there might be overshooting, but we have some bounds on how much. First, since there might be overshooting now, we need to replace $a$ and $b$ with $E(M_\tau\mid \tau = \tau_a)$ and $E(M_\tau\mid \tau=\tau_b),$ and then we get $$ P(M_\tau\ge b) = P(\tau=\tau_b) = \frac{1-E(M_\tau\mid \tau=\tau_a)}{E(M_\tau\mid \tau=\tau_b)-E(M_\tau\mid \tau=\tau_a)}.$$
Then for bounds, we have $$ b\le E(M_\tau\mid \tau=\tau_b) \le \frac{3}{2}b$$ and $$ \frac{a}{2}\le E(M_\tau\mid \tau=\tau_a) \le a.$$
Taking the possibilities that result in the smallest and largest value for $P(M_\tau\ge b)$ gives the bounds in the problem.