A question on CMRR of in-amp and difference amplifier
The outputs of A1 and A2 are out of phase for signals but in phase for common mode errors and the gain of both is the same (often by laser trimming the internal feedback resistors).

A3 therefore acts as a subtraction amplifier for common mode signals; i.e. the outputs of A1 and A2 will be the same for any given common mode input assuming tight matching and therefore A3 will reject this by its own common mode rejection.

Assuming A1 and A2 have tightly coupled performance (on the same silicon they will) then the common mode error will be greatly reduced.

Here is an example:

Assume that all 3 amplifiers have a CMRR of 40dB and all amplifiers are unity gain (for simplicity) and the resistor match is perfect.

Then if the common mode input voltage is 1V (for example), then the output of A1 and A2 (for common mode) is 10mV yielding a 10mV common mode input to A3.

This is then subject to the 40dB CMRR of A3, yielding an output of 100uV of common mode signal.

The overall CMRR is, therefore \$ \frac {V_{in(CMRR)}} {V_{out(CMRR)}} \$ = \$ \frac {1V} {100 uV} \$ = 10,000 = 80dB