acid base - How to create a buffer using CH3COOH and CH3COONa that has a pH of exactly 3.75?

I need to create a buffer using $\ce{CH3COOH}$ and $\ce{CH3COONa}$ that has a pH of exactly $3.75$.

A $\pu{50 mL}$ sample of your buffered solution will have to be able to withstand the addition of $\pu{25.0 mL}$ of $\pu{0.100 M}$ $\ce{NaOH}$ solution.

The buffered solution will break after the addition of no more than $\pu{35.0 mL}$ of the $\pu{0.10 M}$ $\ce{NaOH}$.

Here is what I have so far:

$$\mathrm{pH}= 3.75 \implies [\ce{H+}] = \pu{10^{-3.75} M} = \pu{1.78 \times 10^-4 M}$$

\begin{align} \ce{NaCH3COO &-> Na+ + CH3COO-} &&\text{(common ion effect)}\\ \ce{CH3COOH &<=> H+ + CH3COO-} \end{align}

\begin{align} K_\mathrm{a} &=\frac{[\ce{H+}][\ce{CH3COO-}]} {[\ce{CH3COOH}]}\\ 1.8\times{10^{-5}} &=\frac{(1.78\times{10^{-4}})[\ce{CH3COO-}]} { [\ce{CH3COOH}]}\\ \frac{[\ce{CH3COO-}]}{[\ce{CH3COOH}]} &=\frac{(1.8\times{10^{-5}})}{(1.78\times{10^{-4}})}\\ \frac{[\ce{CH3COO-}]}{[\ce{CH3COOH}]} &= 0.1011 : 1~\text{(ratio)} \end{align}

I found the ratio of $[\ce{NaCH3COO-}]$ to $[\ce{CH3COOH}]$, but I don't know what to do from here.

EDIT: "Break": The point where there's no acid left and the solution stops acting as a buffer. "Withstand": meaning that there can be a $\mathrm{pH}$ change of at most $1.0$. Thanks, guys!