Difference amplifier question
The fastest way to get the answer for me is to use the Extra-Element Theorem or EET (https://en.wikipedia.org/wiki/Extra_element_theorem). If I consider the resistor \$R_G\$ as the extra element, I can set it to infinity for my reference gain and solve the below circuit:



This is a classical differential configuration and by applying superposition you should get:

\$V_{out}=-V_1\frac{2R_2}{R_1}+V_2(\frac{2R_2}{R_1+2R_2}(1+\frac{2R_2}{R_1}))\$ which simplifies into:

\$H_{ref}=\frac{V_{out}}{V_1-V_2}=-2\frac{R_2}{R_1}\$

Then you carry on by determining \$R_d\$ which is the resistance seen from \$R_G\$ terminals when the input voltage (the stimulus) is reduced to 0 V (replace the source by a short circuit). The equivalent circuit is below:



By inspection and considering the component values, the voltage across the test generator is 0 V (very low value in the simulation). In this case, \$R_d=0\;\Omega\$

We need to determine \$R_n\$ which is the resistance seen from \$R_G\$ terminals when the response is nulled while the stimulus is a back in place. The equivalent circuit is below:



Again, by inspection, the resistance seen across \$R_G\$'s terminals in this configuration is the sum of the parallel combinations of \$R_2\$ which finally leads to \$R_n=R_2\$. Applying the EET, we have

\$H=H_{ref}\frac{1+\frac{R_n}{R_G}}{1+\frac{R_d}{R_G}}=-2\frac{R_2}{R_1}(1+\frac{R_2}{R_G})\$

This is the way the EET works, by splitting the problem in different small pieces that you individually solve and assemble at the end. You can see an introduction to the Fast Analytical Circuits Techniques (FACTs) in a seminar taught at APEC in 2016

http://cbasso.pagesperso-orange.fr/Downloads/PPTs/Chris%20Basso%20APEC%20seminar%202016.pdf

and find a lot of examples solved with the FACTs here

http://cbasso.pagesperso-orange.fr/Downloads/Book/List%20of%20FACTs%20examples.pdf