The period of the pendulum is roughly
$T\approx2\pi\sqrt{\frac{L}{F}}$
Where $L$ is its length and $F$ is the downward-pulling force (usually gravity). Let's call our reference period $T_{rest}$.
Now let's examine your cases.
1. When the train is in circular motion in a curve of radius R with constant speed.
If the train is in circular motion the pendulum experiences a fictitious centrifugal force that points radially outward. Firstly, that means that the total force on the pendulum is not straight down any more, but somewhat diagonally downward/outward. Since you have to add this force (vectorially) to the gravitational force, $F$ will also be larger than before. Looking at our equation above, this means that indeed $T_1 < T_{rest}$.
2. The train is going up a constant slope with constant speed.
If the train travels along a straight line (and a constant slope is a straight line) at constant speed, then the pendulum is in an inertial frame. Therefore, there are no fictitious or other forces adding to gravity. Hence, $T_2 = T_{rest}$.
3. The train moves over a hill of radius R with constant speed.
This one is tougher. The centrifugal force is still pointing radially outwards - but now it is not perpendicular to the gravitational force any more. For instance, at the top of the hill, it is pointing upward, and thus cancelling gravity partly. At other positions it will point diagonally upward. This case needs its own somewhat more involved treatment, because the total $F$ is not constant. Therefore, $T_3$ will also change gradually. In particular, it will be maximal at the top of the hill, and will decrease symmetrically down both sides of the hill.
Sorry that I don't have any pictures to clarify this. Feel free to ask if anything is unclear.
You see, when you have a pendulum with friction you account it by including a force $\vec{F}_r=-b\vec{v}$. Then your differential equation for the pendulum is $$ml\ddot{\theta}=-mg\theta-bl\dot{\theta}\iff\ddot{\theta}+\frac{b}{m}\dot{\theta}+\frac{g}{l}\theta=0$$The solution of this differential equation depends on the values of $b$, $m$ and $l$ but since you are asking for a period, we will assume that the pendulum oscilates. Then you've got a solution with angular frequency $$\omega=\sqrt{\frac{g}{l}-\frac{b^2}{4m^2}}$$Therefore, the period of this osscilator is $$T=\frac{2\pi}{\sqrt{\frac{g}{l}-\frac{b^2}{4m^2}}}$$That shows how the period changes compared to the undamped pendulum.
Best Answer
The center of mass of the system consisting of the bob and the mercury inside it shifts downward, effectively increasing the length. However, the final position of the center of mass after all the fluid has drained out is at the center of the bob, so the center of mass shifts up again after some time. Thus, the time period first increases due to the effective increase in length, then decreases since the center of mass shifts back to the center of the bob.