What is the pH at both equivalence points of titration between diprotic tartaric acid and NaOH?
For the 1st equivalence point, the tartaric acid $\ce{H2A}$ is titrated to $\ce{NaHA}$

For the solutions of acid salts, we can with advantage use the formula

$$\mathrm{pH}=\frac{\mathrm{p}K_\mathrm{a1}+\mathrm{p}K_\mathrm{a2}}{2}$$

that can be derived from equilibrium

$$\ce{2 HA- &lt;=&gt; H2A + A^2-}$$

For the 2nd equivalence point there is  $\ce{Na2A}$ solution, which is hydrolyzed with equilibrium

$$\ce{A^2- + H2O &lt;=&gt; HA- + OH-}$$

With neglecting OH- from water hydrolysis, $[\ce{HA^-}] = [\ce{OH^-}]$

$$\begin{align}
K_\mathrm{a2}&amp;=\frac{[\ce{H+}][\ce{A^2-}]}{[\ce{HA-}]}\\
\mathrm{p}K_\mathrm{a2}&amp;=\mathrm{pH} - \log{[\ce{A^2-}]} +\log {[\ce{HA-}]}\\
&amp;=\mathrm{pH} - \log{[\ce{A^2-}]} +\log {[\ce{OH-}]}\\
&amp;=\mathrm{pH} - \log{[\ce{A^2-}]} -\mathrm{pOH}\\
&amp;=\mathrm{pH} - \log{[\ce{A^2-}]} -14 + \mathrm{pH}\\
\mathrm{pH}&amp;=7+\frac12\left(\mathrm{p}K_\mathrm{a2} +\log{ c_\mathrm{\ce{Na2A}}}\right)\\
\end{align}$$

The molarity of $\ce{NaOH}$ solution $c_{\ce{NaOH}}=\frac{m}{M\times V}=\pu{0.09801  mol/L}$

The volume of $\ce{NaOH}$ solution for given volume of tartaric acid solution

$$V_{\ce{NaOH}}=V_{\ce{H2A}}\times \frac{2\times c_{\ce{H2A}}}{c_{\ce{NaOH}}}\\=0.0102\times V_{\ce{H2A}}$$

The final concentration 

$$\begin{align}
[\ce{Na2A}]&amp;=\frac{[\ce{H2A_\mathrm{init}}]\cdot V_{\ce{H2A}}}{V_{\ce{H2A}} +V_{\ce{NaOH}}}\\
&amp;=\frac{[\ce{H2A_\mathrm{init}}]\cdot V_{\ce{H2A}}}{V_{\ce{H2A}} + 
0.0102  \cdot  V_{\ce{H2A}}}\\
&amp;=\frac{[\ce{H2A_\mathrm{init}}]}{1+ 0.0102}\\
&amp;=5\times 10^{-4}/1.0102\\
&amp;=\pu{4.9495\times10^{-4} mol/L}\\
&amp;\overset{\mathrm{rounding}}=\pu{4.95\times 10^{-4} mol/L}\\
\end{align}$$