Solution

1. In the frame of reference of the stations:
Let $L$ be the distance between the stations, $v$ be the velocity of the probe relative to the stations, and $c$ be the speed of light.
The time of flight of the probe from the first to the second station is:
$$t_{\text{probe}} = \frac{L}{v}$$
The light signal makes three trips between the stations (1 $\to$ 2, 2 $\to$ 1, 1 $\to$ 2). Its total travel time is:
$$t_{\text{light}} = \frac{L}{c} + \frac{L}{c} + \frac{L}{c} = \frac{3L}{c}$$
According to the condition, the probe and the signal arrive at the second station simultaneously:
$$t_{\text{probe}} = t_{\text{light}} \implies \frac{L}{v} = \frac{3L}{c}$$
From this, we find the velocity of the probe relative to the stations:
$$v = \frac{c}{3}$$

2. In the frame of reference of the probe:
According to the principle of relativity, the relative velocity of the stations in the probe's frame of reference, $u$, must be equal to $v$. Let's prove this kinematically.
In the probe's frame of reference, both stations move towards it with a velocity $u$. The distance between them undergoes relativistic length contraction:
$$L' = L \sqrt{1 - \frac{u^2}{c^2}}$$
The time it takes for the second station to reach the stationary probe is:
$$t'_{\text{st}} = \frac{L'}{u}$$
Consider the motion of the light signal in this frame of reference. The speed of light is always $c$.

• The signal goes from the first station to the second. The second station moves towards the light with velocity $u$, so the closing speed is $c+u$. Time: $t'_1 = \frac{L'}{c+u}$
• The signal reflects from the second station to the first. The light moves in the same direction as the first station. Their closing speed is $c-u$. Time: $t'_2 = \frac{L'}{c-u}$
• The signal again goes from the first to the second station (towards each other). Time: $t'_3 = \frac{L'}{c+u}$

Total travel time of the light:
$$t'_{\text{light}} = \frac{L'}{c+u} + \frac{L'}{c-u} + \frac{L'}{c+u} = L' \left( \frac{2}{c+u} + \frac{1}{c-u} \right)$$
Equating the travel times of the stations and the light ($t'_{\text{st}} = t'_{\text{light}}$):
$$\frac{L'}{u} = L' \left( \frac{2}{c+u} + \frac{1}{c-u} \right)$$
Canceling $L'$ and bringing to a common denominator:
$$\frac{1}{u} = \frac{2(c-u) + (c+u)}{(c+u)(c-u)}$$
$$\frac{1}{u} = \frac{3c - u}{c^2 - u^2}$$
$$c^2 - u^2 = 3uc - u^2$$
$$c^2 = 3uc \implies u = \frac{c}{3}$$
The velocity of the stations relative to the probe is $\frac{c}{3}$.