From the linked datasheet, this pot is a 2W variety. And on page 26, the "tapers" listed are S-9 and S-10, which are fairly exotic. (A "taper" is how the resistance relates to rotation angle; linear is a linear relation between angle and resistance, log is a logarithmic relation, etc.) This is likely where the large price tag comes from, since most pots are 1/4W or less and have a common linear, log, or anti-log taper.
Now trying to decipher your measurements,
- 0° to 130° = 4Ω.
- 130° to 175° = ~5kΩ *
- 175° to 312° = 4Ω
*Now does this change gradually (and linearly), from 4Ω at 130°, up to 5kΩ at 152°, then back down to 4Ω at 175°?
This would make sense if terminals 1 and 3 were wired together AND the pot were a 10k variety. This is because terminal 2 is the "wiper", and when terminals 1 and 3 are connected, the maximum resistance would be seen at 152.5° - exactly middle of the rotation. Since there isn't one 10k resistance but two (one 10k from terminal 2-1, another 10k from 2-3), two 10k resistors in parallel = 5kΩ.
Now as for "amplifying" this or replacing it with some other device, well that depends on what it's connected to. If it's connected to a motor, then the motor will require more power than a standard 1/4W pot could provide so that would literally burn up. Such a pot can be used to control a motor but this is not done directly; a driver circuit must be used to take the weak currents coming out of the pot and "amplifying" them to drive the motor. The construction of such a device is rather involved, and the specs of the motor and power source must be known.