pH calculations - please help?

1)
Refer to: http://www4.ncsu.edu/~franzen/public_html/CH201/data/Acid_Base_Table.pdf
Ka for acetic acid = 1.8 × 10⁻⁵
(Different results would be obtained when different values of Ka are used.)

Equation for the reaction :
CH₃COOH(aq) ⇌ CH₃COO⁻(aq) + H⁺(aq) …… Ka = 1.8 × 10⁻⁵

pH = pKa + log([CH₃COO⁻]/[ CH₃COOH]) = -log(1.8 × 10⁻⁵) + log(0.5/0.5) = 4.7


2.
As Ka for CH₃COOH is very small and due to the common ion effect in the presence of CH₃COO⁻ ions, the dissociation of CH₃COOH is negligible.
No. of moles of CH₃COOH = (0.05 mol/L) × (5/1000 L) = 2.5 × 10⁻⁴ mol

You need to know:
Ka of CH3COOH = 1.8*10^-5
pKa = -log Ka = -log (1.8*10^-5) = 4.74
If [CH3COOH] = [CH3COONa] , as in this case , then
pH = pKa
pH = 4.74

You can solve this by using the Henderson - Hasselbalch equation
pH = pKa + log ([CH3COONa]/[CH3COOH]
pH = 4.74 + log ( 0.05/0.05)
pH = 4.74 + log 1.0
pH = 474 + 0
pH = 4.74

Is this exactly what the question asked?
Calculate the number of moles of acetic acid present:
In a 0.05M solution of CH3COOH ,
1000mL contains 0.05 mol CH3COOH
5mL contains 5/1000*0.05 = 0.00025 mol CH3COOH
The CH3COONa does not change the concentration of CH3COOH.