How to calculate gain of wheatstone bridge+differential amplifier?
A) Create a Wheatstone bridge by Circuitlab or by hand and balance it with a resistor of 330 Ohm and a voltage source (DC) of 5V for 23 degrees! Write down calculations. 


Providing a complete solution for a study or school project, does not use anything in learning the specific topic, that's why I will just provide a explanation of the resistor circuit. The rest about the Differential Amplifier can be found on many internet resources. 

I've recreated the shown circuit to show the correct annotations. 



R0 is the resistor connected in Series with the RTD Resistor Rsensor. Given the formula for Rsensor:

Rsensor = R0 x (1+ a Delta T) translates to Rsensor = 330 Ohms *(1+0), which is 330 Ohms since temperature has not changed.

R0 and Rsensor build a Voltage Divider. By building the exact same voltage divider and connecting both middle points to a differential opamp, you have a differential readout because of:

Vout 1 = (5V X Rsensor) / (R0 + Rsensor) =  (5V x 330) / (330 + 330)  = 2.5V

same formula for the second divider:

Vout 2 = (5V X R2) / (R1 + R2) =  (5V x 330) / (330 + 330)  = 2.5V

The differential amplifier will output the amplification of the difference between the both input terminals which would be : 2.5V-2.5V = 0V

When the temperature changes, Rsensor will change its value, for example become 470Ohm:

Vout 1 = (5V X Rsensor) / (R0 + Rsensor) =  (5V x 470) / (330 + 470)  = 2.93V

R1 and R2 would stay the same:

Vout 2 = (5V X R2) / (R1 + R2) =  (5V x 330) / (330 + 330)  = 2.5V

Which would result in the following difference: 2.93 - 2.5 = 0.43V

Rsensor will change 0.33% for each Kelvin difference from 23°C. In terms of change 1Kelvin = 1 Celsius.

For example:
Temperature rises by 10 Kelvin  which would result in:

Rsensor = 330Ohm x (1+ (0.0033 * 10)) = 340.89 Ohm

Hopefully this will get you further with your task!