NaOH is a strong base therefore complete dissociation takes place.
number of NaOH moles dissolved - 0.36 g/ 40 g/mol = 0.009 mol
Concentration of NaOH - 0.009 mol / 0.5 L = 0.018 M
molar ratio of NaOH to OH⁻ is 1:1, therefore [NaOH] = [OH⁻]
pOH = -log [OH⁻]
pOH = -log (0.018)
pOH = 1.74
pH + pOH = 14
pH = 14 - 1.74
pH = 12.26