Effect on mass when decreasing pressure in equilibrium reaction
This is an approach to Question 2.
The reaction equilibrium condition for the system is:  $$\frac{V}{RT}\frac{n_C}{n_A n_B}=K_P\tag{1}$$Let $n_{A0}$, $n_{B0}$, and $n_{C0}$ be the number of moles at equilibrium for each species when the volume is equal to $V_0$.  Then we have
$$\frac{V_0}{RT}\frac{n_{C0}}{n_{A0} n_{B0}}=K_P\tag{2}$$.  For a small change in volume dV, we can write the new number of moles of A, B, and C at equilibrium as
$$n_A=n_{A0}-dn\tag{3a}$$
$$n_B=n_{B0}-dn\tag{3b}$$
$$n_C=n_{C0}+dn\tag{3c}$$Combining Eqns. 1-3 yields:$$\left(1+\frac{dV}{V_0}\right)=\frac{(1-dn/n_{A0})(1-dn/n_{B0})}{(1+dn/n_{C0})}$$And, linearizing this equation with respect to dn yields:$$\frac{dV}{V_0}=-\left(\frac{1}{n_{A0}}+\frac{1}{n_{B0}}+\frac{1}{n_{C0}}\right)dn$$This equation indicates that if dV is negative, then dn is positive, and the reactants decrease while the product increases.
If we apply this same methodology to the concentration of A, we obtain
$$dC_A=-\frac{n_{A)}}{V_0}\left(\frac{dn}{n_{A0}}+\frac{dV}{V_0}\right)$$Combining this with the previous equation then yields:
$$\frac{dC_A}{dV}=-\frac{n_{A0}}{V_0^2}\left[\frac{1/n_{B0}+1/n_{C0}}{1/n_{A0}+1/n_{B0}+1/n_{C0}}\right]$$The term in brackets is always positive, so, when the volume decreases, the concentration of A increases at all points along the equilibrium contour.  This conclusion is independent of the specific value of the equilibrium constant.