How to calculate partition coefficients?
Equilibrium are really dependent on the chemical activities of the species not their concentrations. So an overriding assumption for both parts is that the activities of the chemical species is the same as the concentration. That is a reasonable for these solutions. As a rule of thumb, for solutions with concentrations more than 0.1 molar the assumption is dicey. 

I think you're very close to the right solution for part 1, but I'd state it a bit differently.


Since pH is given, $\ce{[H^+]}$ in the aqueous phase is determined by



  
    $[\ce{H^+}] = 10^{-\mathrm{pH}} = 10^{-3.21} = 6.2 \cdot 10^{-4}~\mathrm{mol\,dm^{-3}}$ 
    
    
      The mantissa of the pH only has two significant figures, so should the $\ce{[H^+]}$
    
  



We know that in the aqueous layer the charges have to balance so $\ce{[H^+] = [A^-] + [OH^-]}$, but $\ce{[A^-] &gt;&gt; [OH^-]}$ so we can assume that $\ce{[H^+] = [A^-]}$
Using $\ce{[H^+] = [A^-]}$ the $K_\mathrm{a}$ equation can be solved for the $\ce{[HA]}$ present in aqueous layer which gives $0.038~\mathrm{mol\,  dm^{-3}}$.


$$\ce{[HA]} = \dfrac{\ce{[H^+][A^-]}}{1\times10^{-5}} = \dfrac{(6.2\times10^{-4})^2}{1\times10^{-5}} = 0.038$$


Next, the $\ce{[A-]}$ in the aqueous phase is insignificant compared to the $\ce{[HA]}$ in the aqueous phase, so the moles of $\ce{HA}$ present in the organic layer can be determined by subtracting the moles of $\ce{HA}$ present in aqueous layer (0.038*0.5=0.019)from the initial moles of $\ce{HA}$ (0.057*0.50 = 0.029), which results in a final concentration of $0.020~\mathrm{mol\,  dm^{-3}}$ of $\ce{HA}$ in the organic layer.
Finally, $K_\mathrm{D} =
   \dfrac{[\ce{HA}_\mathrm{aqueous~layer}]}{[\ce{HA}_\mathrm{organic~layer}]}
   = \dfrac{0.038~\mathrm{mol\,  dm^{-3}}}{0.020~\mathrm{mol\,  dm^{-3}}} = 1.9$



  
    The $\ce{[HA]}$ in both layers is known to two significant figures so the $K_D$ value should have two significant figures too.
  




For Question 2.


The reaction between $\ce{HA}$ and $\ce{NaOH}$ in aqueous solution is, as you noted,


$$\ce{HA + NaOH -&gt; Na^+ + A^- + H2O}$$


Assume none of $\ce{NaA}$ migrates into the organic phase. 
Assume that no $\ce{NaA}$ forms in the aquaeous phase either, which is to say that both the $\ce{Na^+}$ and $\ce{A^-}$ ions are completely solvated. 
Since $\ce{[Na^+]}= 0.027$ and the volume is 0.5000 $\mathrm{dm}^{-3}$, the number of moles of $\ce{NaOH}$ present is $0.0135$.
Started with 0.0295 moles of HA total, so the solution will be acidic. 
Since the charges must balance $\ce{[Na^+] + [H^+] = [A^-] + [OH^-] }$, but $\ce{[A^-] &gt;&gt; [OH^-] }$ and $\ce{[Na^+] &gt;&gt; [H^+]}$ so $\ce{[Na^+] \approx [A^-] }$ which means that there are $0.0135$ moles of $\ce{[A^-]}$ in the aqueous solution, and 0.016 moles of $\ce{HA}$ split between the aqueous phase and the organic phase.  
From above $0.016 = \ce{0.5[HA]_{aq} + 0.5[HA]_{org}}$ and using $K_D$ we can determine $\ce{[HA]_{aq} = 0.021}$ and $\ce{[HA]_{org} = 0.011}$
We can use the $K_a$ to solve for $\ce{[H^+]}$


\begin{align}
1\times10^{-5} &amp;= \dfrac{\ce{[H^+][A-]}}{\ce{[HA]}}\\
\ce{[H^+]} &amp;= \dfrac{1\times 10^{-5}\ce{[HA]}}{\ce{[A^-]}} = \dfrac{(1\times 10^{-5})(0.021)}{0.027} = 7.7\times10^{-6}\\
\mathrm{pH} &amp;= 5.11
\end{align}