Weight of sodium acetate required to get to certain pH
It's not really a matter of remembering a certain formula, this is not a conceptual thinking and should be avoided. When approaching these types of questions, always start with a balanced chemical equation. In this case we deal with a salt formed by a weak acid and a strong base, which hydrolyses according to the following net ionic equation:

$$\ce{CH3COO- + H2O &lt;=&gt;[$K_\mathrm{h}$] CH3COOH + OH-}$$

$$
\begin{align}
K_\mathrm{h} &amp;= \left.\frac{[\ce{CH3COOH}][\ce{OH-}]}{[\ce{CH3COO-}]}\qquad\right|\cdot\frac{[\ce{H+}]}{[\ce{H+}]}\\
&amp;= \frac{[\ce{CH3COOH}][\ce{OH-}][\ce{H+}]}{[\ce{CH3COO-}][\ce{H+}]}\\
&amp;= \frac{K_\mathrm{w}}{K_\mathrm{a}}\label{eq:1}\tag{1}
\end{align}
$$

At the same time disregarding autoprotolysis of water, and taking into account that $[\ce{CH3COOH}]\approx [\ce{OH-}]$ and $[\ce{CH3COO-}]\approx c$ ($c$ is the concentration of sodium acetate), hydrolysis constant is

$$K_\mathrm{h} = \frac{[\ce{OH-}]^2}{c}\label{eq:2}\tag{2}$$

Knowing that

$$[\ce{OH-}] = \frac{K_\mathrm{w}}{[\ce{H+}]} = \frac{K_\mathrm{w}}{10^{-\mathrm{pH}}}\label{eq:3}\tag{3}$$

one can equate \eqref{eq:1} and \eqref{eq:2}, substitute $[\ce{OH-}]$ with \eqref{eq:3}, and express concentration as

$$
\require{cancel}
c = \frac{K_\mathrm{a}[\ce{OH-}]^2}{K_\mathrm{w}} = \frac{K_\mathrm{a}K_\mathrm{w}^{\cancel{2}1}}{10^{-2\mathrm{pH}}\cancel{K_\mathrm{w}}} = \frac{K_\mathrm{a}K_\mathrm{w}}{10^{-2\mathrm{pH}}}\tag{4}
$$

Finally, mass $m$ of sodium acetate can be found from its molar mass $M$ and volume of the solution $V$:

$$m = cVM = \frac{VMK_\mathrm{a}K_\mathrm{w}}{10^{-2\mathrm{pH}}} = \frac{\pu{2.50 L}\cdot\pu{82.03 g mol-1}\cdot 1.75\cdot 10^{-5} \cdot 10^{-14}}{\pu{10^{-2\cdot 9} mol-1 L}} = \pu{35.89 g}\tag{5}$$