Question 20.21 and 23

20)
CuSO4.5H2O(s)  <---> CUSO4(s)  + 5H2O
kp = P(eq)H2^5
P gas (H2) = 1(0^-10)^1/5 = 0.01 atm
use ideal gas law
PV = nRT
get n of water = RT/PV = 1*10^-3 moles
It means  1*10^-3 moles /5 moles of CuSO4.5H2O(s) has undergone dissociation
so left amount CuSO4.5H2O(s)= 0.01 - 0.0002 = 9.8*10^-3 moles
amount of CuSO4 (s) = moles of H2O / 5 = 1*10^-3 moles /5 = 0.0002 moles CuSO4(s)

Q21
A part is wrong because Delta G = -RTlnKp
and Kp in this case = P(H2O)^2
B) it is correct because  P(h2O0 = 15.2 torr is actually  equilibrium pressure of water .
(15.2/760) ^2 = Kp = 4.0*10^-4
It means no more water can be formed as it is already 100% saturated with water vapours.
C )If you convert 24 torr into atm and take square of it .you will get a value of Qp which is more than Kp .Means pressure of water is already more than equilibrium pressure and hence equilibrium shifts to left.
D) In open container and dry atmosphere ,equilibrium  will shift more to right   to get equilibrium pressure.However container is open so it will not be  able to achieve equilibrium and  water will dehydrate from CuSO4.5H2O(s) to give CuSO4.3H2O(s)

In 23
all are correct  as we are discussing concentration of HI ...so any factor which changes K  equilibrium constant and any factor which changes volume can change the concentration of HI

in q23 for D part how does concentration remain same.

You can set up equilibrium from any side
---either from 2 moles of HI or 2 moles of the reactants ( 1 mole H2 and 1 mole I2 ) .You will get same concentration at equilibrium if it is done in the same vessel.