Question 1
[tex]\text{pH}=-\log(\text{H}^{+})\\\\13=-\log(\text{H}^{+})\\ \\ \text{[H}^{+}]}=1 \times \boxed{10^{-13} \text{ moles per liter}} \\ \\ \text{[H}^{+}] \times \text{[OH}^{-}] =1 \times 10^{-14} \text{ M}\\\\\text{[OH}^{-}]=\boxed{1 \times 10^{-1} \text{ moles per liter}}[/tex]
Question 2
[tex]\text{pH}=-\log(\text{H}^{+})\\\\\text{pH}=-\log(1.0 \times 10^{-4})=\boxed{4}[/tex]