Consider an amphoteric hydroxide, m(oh)2(s), where m is a generic metal. estimate the solubility of m(oh)2 in a solution buffered at ph

Missing data in your question: (please check the attached photo)
from this balanced equation:
M(OH)2(s) ↔ M2+(aq) + 2OH-(aq) and when we have Ksp = 2x10^-16
∴Ksp = [M2+][OH]^2
2x10^-16 = [M2+][OH]^2
a) SO at PH = 7 
∴POH = 14-PH = 14- 7 = 7
when POH = -㏒[OH]
7= -㏒[OH]
∴[OH] = 1x10^-7 m by substitution with this value in the Ksp formula,
∴[M2+] =Ksp /[OH]^2
            = (2x10^-16)/(1x10^-7)^2
             = 0.02 M
b) at PH =10
when POH = 14- PH = 14-10 = 4 
when POH = -㏒[OH-]
            4  = -㏒[OH-]
∴[OH] = 1x10^-4 ,by substitution with this value in the Ksp formula
[M2+] = Ksp/ [OH]^2
          = 2x10^-16 / (1x10^-4)^2
          = 2x10^-8 M
c) at PH= 14 
when POH = 14-PH
                   = 14 - 14 
                   = 0
when POH = -㏒[OH]
              0 = - ㏒[OH]
∴[OH] = 1 m 
by substitution with this value in Ksp formula :
[M2+] = Ksp / [OH]^2
          = (2x10^-16) / 1^2
          = 2x10^-16 M

Must be negative.substance increases