Numerical on acetic acid - sodium acetate buffer

Starting with 1 L of 2.0 M $\ce{CH3COOH}$, we wish to make a buffer solution of pH=4.00.

Consider two ways to make the buffer: a) One way would be to add sodium acetate. How many moles must be added?(assume no volume change) b) Another way would be to add NaOH and have it react with acetic acid to produce the correct acetate ion concentration. How many moles of NaOH do we need?

Part a) - My work:

$$\ce{CH3COOH <=> H+ + CH3COO-}$$

From Henderson’s equation, $$\ce{pH = pK_a+ log \frac{[base]}{[acid]}}$$

$\ce{pK_a}$ of acetic acid = 4.74

Plugging in, I get: 4= 4.74 + log$\ce{\frac{[CH3COO-]}{[2]}}$

For my final answer, I got $\ce{[CH3COO-]}$ = 0.364 M. So 0.364 moles of $\ce{CH3COO-}$ need to be added to 1L of solution.

Part b)-My work:

$$\ce{CH3COOH + OH- <=> H2O + CH3COO-}$$

$\ce{K_a = \frac{[CH3COOH]}{[OH-]}}$

I have no idea where to go from here. Any help would be greatly appreciated!

Suppose you add $x$ moles of NaOH, So, you acetate ion will form in $x$ moles. NaOH will be totally consumed and acetic acid will be left $2-x$ moles. Now you solve the Henderson-Haselbach equation as, $$\ce{4=4.74 + log(x/ (2 - x))}$$ You will get $x$ as $0.308$ moles, which is your solution.