The pH of a dissolution of NaOH is 13, which Volume of water do I have to add to make the pH 12?

You need to write down the relevant equations.

NaOH \rightarrow Na^+ + OH^-

H_2O \rightleftharpoons H^+ + OH^-

You know the pH of the NaOH, you can find the concentration of OH^- in the solution.

Remember that:

pH + pOH = 14

pH = -\log[H^+]

pOH = -\log [OH^-]

K_w = 1.0 \times 10^{-14} = [H^+][OH^-]

Can you give your problem a try now? :)