Answer
a) $1.34\dot{}10^{-5}\ M$
b) $4.45\dot{}10^{-2}\ M$
Work Step by Step
a) $Ksp=[Ag^+][Cl^-]=s^2$
$s=\sqrt{1.8\dot{}10^{-10}}=1.34\dot{}10^{-5}\ M$
b) $Knet=Kf\dot{}Ksp=[Ag(NH_3)_2^+][Cl^-]/[NH_3]^2$
Assuming constant [NH3]:
$1.98\dot{}10^{-3}=s^2/1.0^2$
$s=4.45\dot{}10^{-2}\ M$