What is the maximum and minimum input voltage in millivolts that can be amplified?
According to the schematic, the module consists of two cascade-connected inverting amplifiers, each of them has a gain of 10. So the maximum amplification can be 10 x 10 = 100.
The module accepts single-supply, the input is AC coupled (i.e. can't be used with DC signal amplification) and the input will be offset by half the supply voltage. The output of an LM358 can swing up to supply voltage minus 1.5 Volts, so we can write the following equation:
$$
\mathrm{V_i\cdot 100+\frac{V_{CC}}{2}\leq V_{CC}-1.5 \\
\Rightarrow V_i \leq \frac{V_{CC}-3}{200}
}
$$
Where VCC is the supply voltage, and Vi is the peak of the input voltage.
For 5 VDC supply, the input's peak shouldn't exceed 10 mV i.e. you shouldn't apply 20 mVpp or else the output will be clipped/saturated. The input voltage range can be increased by supplying the module with higher voltage (e.g. 12 VDC or even 24 VDC).
The amplification can be changed using the screw on the module. Is there a way to set the amplification to an exact value?
Hard to tell. The pot seems to be a multi-turn one so you can't estimate easily without measuring.
Note that I didn't take the offset error of the LM358 into account. For any op-amp-based amplifier, the offset error will be amplified and reflected to the output. The typical offset error of LM358 is 2 mV and can be as high as 3 mV. This means that if your sensor's output (i.e. the signal to be amplified) is less than or equal to 2 mV then you will highly likely get a huge measurement error. This can be rectified by running a zero-calibration, though. But if the input voltage is in mV range (e.g. less than 5 mV) then you should use an op amp with very low offset error.