How do you calculate the pH of a buffer solution after adding a base?

Use the Henderson-Hasselbalch equation:

pH = pKa - log ( [A-] / [HA])

For the first half of the question (discounting the volume of the Na acetate and assuming the sodium acetate is anhydrous):
 MW of Na acetate (CH3COO Na)  = (12.0 x 2) + (1.00 x 3) + (23.0 x 1) + (16.0 x 2) = 82.0 g/mol

4.95 g Na acetate / (82 g/mol) = 0.0604 mol Na acetate

0.0604 mol Na acetate/0.250 L solution = 0.242 M Na acetate

pH = pKa - (log ( [A-] / [HA]) = (-log (1.8E-5) - log (0.242 M / 0.150 M) = 4.54

For the second part:

MW of NaOH = (23.0 x 1) + (16.0 x 1) + (1.00 x 1) = 40.0 g/mol

0.082 g NaOH /40.0 g/mol = 2.05E-4 mol NaOH

2.05E-4 mol NaOH / 0.010 L = 0.0205 M  NaOH

 pH = pKa - (log ( [A-] / [HA]) = (-log (1.8E-5) - log ( (0.242 M + 0.0205 M) +  / (0.150 M - 0.0205 M) ) = 4.745 - log (0.2625/0.125) = 4.42

First, use the Henderson-Hasselbalch (HH) equation: pH=pKa log([A-]/[HA])
for the first solution this gives pH=4.744+log(.2414/.150)= 4.95

To calculate the addition of NaOH assume that it reacts 100% with the acetic acid and therefore merely subtract the moles NaOH from acetic acid, then use the HH eq. again.
this gives .035449 moles acetic acid in 250 mL solution= .1418M
also, this adds to the amount of sodium acetate by the same amount, so we add that to the equation resulting in the new NaAc concentration of .2495M Na acetate.
using HH we get: pH=4.744+log(.2495/.1418)= 4.99
This also shows the effectiveness of even dilute buffers  (.04 change in pH when adding a strong base.