how do you prepare 100 mL of 0.200 M acetate buffer, pH5.00, starting with pure liquid acetic acid and solutions containing~3 M HCl and ~3 M NaOH

Answer 1

CH3COOH + NaOH ==> CH3COONa + H2O
CH3COOH is the acid; i.e., a
CH3COONa is the base; i.e., b
pKa for acetic acid is pKa = -log Ka. I think pKa for acetic acid is 4.75, at least close to that, but you should confirm that.
pH = pKa + log (b/a)]
5.00 = 4.75 + log b/a.
Solve for the ratio of b/a.That's equation 1.
Equation 2 is a + b = 0.200 M
Solve the eqn 1 and eqn 2 simultaneouly to give values for both acid and base, convert those to volume of acid and base. You probably don't have enough information about the acid such as density or molarity. Also, I note that the HCl and NaOH are "about" 3 M. In practice one calculates the volumes, mixes everything together, measures the pH with a pH meter, then adds HCl or NaOH, drop by drop with stirring until the pH meters reads 5.00.
Post your work if you get stuck with the calculation part.