how to detect variations of small resistances
If the resistance change is indeed so small (which I doubt; it's only 0.25 %), then you need a Wheatstone bridge:  



The amplifier is an instrumentation amplifier, so the feedback is internal. Rx is your variable resistor, and R3 is a fixed 318 &Omega; resistor. R1 and R2 are equal, their actual values are not that important, let's pick 10 k&Omega; for them. Then the voltage at the inverting input of the instrumentation amplifier will be 2.5 V, while the voltage on the non-inverting input will vary between 2.5 V (Rx = 318 &Omega;) and 2.497 V (Rx = 318.8 &Omega;). That's only a 3.14 mV difference, and you'll need a 1000 \$\times\$ gain to get 0 V to 3.14 V out of this. 

The 0.25 % variation is smaller than the tolerance of a 1 % resistor, and even 0.1 % resistors will give you a noticeable offset. If you pick 0.1 % resistors for R1 and R2 then the theoretical 2.5 V may deviate by as much as 2.5 mV, that's 80 % of your full range. So you'll have to trim the resistors so that the output is 0 V at a 318 &Omega; resistance.  



If you use 100 k&Omega; resistors for R5 and R6 and a 10 k&Omega; multiturn for the potmeter you'll be able to adjust the reference voltage between 2.48 V and 2.52 V.