Answer
See the answer below.
Work Step by Step
First equilibrium:
$K_{b1}=[N_2H_5^+][OH^-]/[N_2H_4]$
$8.5\cdot10^{-7}=x^2/(0.010\ M-x)$
$x=9.18\cdot10^{-5}\ M$
Second equilibrium:
$K_{b2}=[N_2H_6^{2+}][OH^-]/[N_2H_5^+]$
$8.9\cdot10^{-16}=y(9.18\cdot10^{-5}+y)/(9.18\cdot10^{-5}-y)$
$y=8.9\cdot10^{-16}\ M$
a) $[N_2H_6^{2+}]=8.9\cdot10^{-16}\ M, [OH^-]=[N_2H_5^+]=9.18\cdot10^{-5}\ M$
b) $pH=14+log(9.18\cdot10^{-5})=9.96$
