Question

Can someone how to do these problems!!

Answer 1

You can use the follwing equations.
pH=14-pOH
pOH=-log[OH⁻]

a)  pOH=-log10⁻¹²
     pOH=12
     pH=14-12
     pH=2

You can also use these equations:
[H⁺]=K(w)/[OH⁻]    (K(w)=1.01×10⁻¹⁴)
pH=-log[H⁺]

b) [H⁺]=(1.01×10⁻¹⁴)/10⁻²M
    [H⁺]=1.01×10⁻¹²M²
    pH=-log(1.01×10⁻¹²)
    pH=12

You can use either method.  It does not really matter.
I hope this helps.  Let me know if anything is unclear and when you do the calculations for c you should get pH=7.