pH of a solution of pure water and sodium hydroxide.

To calculate pH you need to have the concentration of H+ ions in the solution and you use Kw to find this.

Kw=[H+][OH-] = 4.02 x 10^-14 at 318K (from the question)
You have calculated [OH-] in part (iii)
so you have Kw and [OH-] you now rearrange to get [H+] and then use
pH = -log [H+]

Yes, but it's the adding the pure water part that I couldn't figure out.

I just tried it by working out the new concentration after the water was added and I got 10.40 as the pH. Does that sound about right?

oops I missed that bit out - sorry.
I meant to say you have the moles of OH- from (iii) so you need to calc concentration after the dilution.
The moles of OH- stay the same so to calc conc you use the new volume (2 cm3 + 998cm3) and use conc = mol/dm3

Then you have [OH-] and you can put it in the equation and rearrange for [H+]

and yes you have the correct answer.