1. The law of motion of the first tourist: $S=\sqrt{1+6 t}-1$, and the second: $-S=$ $6\left(t-\frac{1}{6}\right)$ for $t \geq \frac{1}{6} ; S=0$ for $t<\frac{1}{6}$.

The required condition is obviously met when both tourists are "on the same side" of the sign. And it is not met when the sign is between them. Therefore, since the first tourist reaches the sign at time $t_{1}$: $\sqrt{1+6 t_{1}}-1=2$, i.e., at $t_{1}=\frac{4}{3}$, and the second - at time $t_{2}: 6\left(t_{2}-\frac{1}{6}\right)=2$, i.e., at $t_{2}=\frac{1}{2}$, then the suitable moments of time are $t \in\left[0, \frac{1}{2}\right] \cup\left[\frac{4}{3},+\infty\right]$. Whether the second tourist is moving or still on a break - does not matter.
Answer: $t \in\left[0, \frac{1}{2}\right] \cup\left[\frac{4}{3},+\infty\right)$.