Kb = (NH4^+)(OH^-)/(NH3)
(NH4^+) = x
(OH^-) = x
(NH4) = 1.0M
Solve for x = (OH^-)

b) Same set up but (NH4^+) = x + 1.0
(OH^-) = x
(NH3) = 1.0-x
This will be a quadratic but you can simplify it by assuming x is so small as to be negligible so x+1.0 = 1.0 and 1.0-x = 1.0. I don't think this will make much difference in the answer but it makes it a bit easier to solve.
Of if you've had the Henderson-Hasselbalch equation you may use that since the NH4Cl/NH3 is a buffered solution.