Answer
(a) The concentration of $OH^-$ produced naturally by water is equal to $1.0 \times 10^{-7}$ which is much smaller than $0.060$ M, and will not affect the number at this amount of significant figures, so we just do not consider it.
(b) Considering significant figures:
$1.0 \times 10^{-7} M + 0.060 M = 0.0000001 M + 0.060 M = 0.060 M $
Work Step by Step
NaOH is a strong base, therefore, 0.060 M of it will produce 0.060 M of $OH^-$.
To calculate the total amount of $OH^-$, we sum the amount given by water, and the amount given by NaOH:
$$[OH^-] = [OH^-]_{H_2O} + [OH^-]_{NaOH}$$
