Along the equator, a train A travels from east to west while a...

Let $v$ be the speed of the trains relative to the Earth, $\omega$ be the angular velocity of the Earth, and $R$ be the radius of the Earth.

The Earth rotates from west to east. The absolute speed of a point on the equator is $\omega R$ from west to east.

Train A travels from east to west. Its absolute speed is $v_A = \omega R - v$.

Train B travels from west to east. Its absolute speed is $v_B = \omega R + v$.

The normal force $N$ exerted by the track on the train provides the necessary centripetal force along with gravity. The equation of motion is $mg - N = \dfrac{m v_{abs}^2}{R}$, which gives N = m \left( g -\dfrac{v_{abs}^2}{R} \right).

For train A, $F_1 = m \left( g - \dfrac{(\omega R - v)^2}{R} \right)$.

For train B, F_2 = m \left( g -\dfrac{(\omega R + v)^2}{R} \right).

Since $(\omega R + v)^2 > (\omega R - v)^2$, it follows that $F_1 > F_2.$