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.
Since your pendulum oscillates about a new angle, call it $\theta_0$, your Taylor series approximation should be about $\theta_0$. So,
\begin{align*}
l \ddot{\theta} & = -g \sin{\theta} + A \cos{\theta} \\
& \approx -g ( \sin{\theta_0} + (\theta - \theta_0) \cos{\theta_0} ) + A ( \cos{\theta_0} - (\theta - \theta_0) \sin{\theta_0}) \text{ for small deviations from } \theta_0 \\
& = A \cos{\theta_0} - g \sin{\theta_0} + \theta_0 (g \cos{\theta_0} + A \sin{\theta_0}) - \theta (g \cos{\theta_0} + A \sin{\theta_0})
\end{align*}
Solving that should give you the period you're looking for.
Best Answer
If you are in the elevator that is accelerating down, but you don't know that; instead you are told you are on a different planet and you have to determine the local acceleration of gravity. Then you would measure the force on a known mass and conclude $g'= F/m$ .
The situation is completely indistinguishable from the one where you are in the accelerating elevator. In other words - you are in a "world" where the restoring force on a pendulum is $mg'\sin\theta$. The derivation of the period then immediately follows along the same lines as it did for the usual situation (where $g'=g$)