Operational Amplifier Problem
This is easier than it looks.  I won't solve it all the way, but lets call the voltage at the top of the pot \$V_a\$ and the voltage at the bottom of the pot, \$V_b\$.  Lets call the positive input \$V_{1+}\$ and the negative, \$V_{1+}\$.  The input terminals of the op amp are \$V_-\$ and \$V_+\$.

Using the rule that no current enters the op amp, it is straightforward to show that \$V_a=2V_- - V_{1+}\$.  Using the rule the \$V_- = V_+\$, it is straightforward to show that \$V_b = 2V_- - V_{1-}\$.  Now, we know the voltage at either side of the pot, and the value of the pot, so \$i_{pot} = \frac{(V_a-V_b)}{(1-X)R} = \frac{V_{1-} - V_{1+}}{(1-X)R}\$

Application of Kirchoff's current law to the node at the top of the pot should give you \$V_2\$