How many grams of sodium phosphate must be added to precipitate as much of one ion as possible?
In an answer that is now hidden user Maurice pointed out the complexities of the chemistry. So let's assume:


All Al present in solution as $\ce{Al^{3+}}$ and any hydoxy complexes or precipitates will be ignored.
All Ca present in solution as $\ce{Ca^{2+}}$ and any hydoxy complexes will be ignored.
All $\ce{PO4^{3-}}$ added to solution stays as that ion and doesn't hydrolyze to $\ce{HPO4^{2-}}$
Two salts that will be produced are $\ce{AlPO4}$ and $\ce{Ca3(PO4)2}$
$K_\mathrm{sp} (\ce{AlPO4})=9.83\times10^{-21}$ and $K_\mathrm{sp} (\ce{Ca3(PO4)2})=2.07\times 10^{-33}$


Concentration of $[\ce{PO4^3-}]$ that will cause $\ce{Al^{3+}}$ to precipitate:
$$[\ce{Al^3+}]=\dfrac{\pu{0.002 mol}}{\pu{200 mL}}=0.01M$$
$$[\ce{PO4^3-}]_\ce{Al}=\dfrac{K_\mathrm{sp}}{\ce{[Al^{3+}]}} = \dfrac{9.83\times 10^{-21}}{\pu{0.01M}}=9.83\times 10^{-19}$$

Concentration of $[\ce{PO4^3-}]$ that will cause $\ce{Ca^{2+}}$ to precipitate:
$$[\ce{Ca^2+}]=\dfrac{\pu{0.001 mol}}{\pu{200 mL}}=0.005M$$
$$[\ce{PO4^3-}]_\ce{Ca}=\sqrt{\dfrac{K_\mathrm{sp}}{\ce{[Ca^{2+}]^3}}} = \sqrt{\dfrac{2.07\times 10^{-33}}{(\pu{0.005 M})^3}}= 1.2890\times 10^{-13}$$

$\therefore \ce{AlPO4}$ will precipitate first. 

The molecular mass of $\ce{Na3PO4}$ is 163.941 g/mol.

Note here that you can't just assume a stoichiometric amount of $\ce{Na3PO4}$ to say three significant figures. The solubility of the $\ce{Ca3(PO4)2}$ is so low that the $\ce{Na3PO4}$ would have to be weighed to an absolutely unrealistic precision.

So we'll go father down fantasy lane and suppose that:


There are $0.02\bar0$ moles of  $\ce{Al^3+}$ 
That we are using God's balance which can weigh an infinitely precise amount of $\ce{Na3PO4}$. 
That the molecular mass of $\ce{Na3PO4}$ is exactly 163.941 g/mol (this is so precise that isotopic ratios limit the value...)
$K_\mathrm{sp} (\ce{AlPO4})=9.83\bar0\times10^{-21}$ and $K_\mathrm{sp} (\ce{Ca3(PO4)2})=2.07\bar0\times 10^{-33}$
That the volume changes in the solution when adding $\ce{Na3PO4}$ or precipitating $\ce{AlPO4}$ are insignificant. 


So if the $[\ce{PO4^3-}] = 1.\bar0\times 10^{-13}$ then no $\ce{Ca3(PO4)2}$ will precipitate. At that concentration the amount of $\ce{Al^3+}$ left in solution will be:

$$\ce{[Al^{3+}]}=\dfrac{K_\mathrm{sp}}{\ce{[PO4^{3-}]}} = \dfrac{9.83\times 10^{-21}}{1.\bar0 \times \pu{10^{-13} M}} =9.83\times 10^{-8}$$

So $\pu{0.001999998034000 mol}$ of $\ce{AlPO4}$ will precipitate. 

For 200 ml of a $1.\bar0 \times \pu{10^{-13} M}$ solution there are $2.\bar0 \times \pu{10^{-14} moles}$

So the total moles of $\ce{Na3PO4}$ needed is $\pu{0.001999998034020  mol}$ or ‭$\pu{0.327881677695273 g}$.

% Al left in solution $= \dfrac{0.002\bar0 - \pu{0.001999998034000 mol}}{0.002\bar0} =‭‭0.0000983‬\% $

Again the problem requires such a precise amount of $\ce{PO4^{3-}}$ to precipitate "all" the $\ce{Al^{3+}}$ yet leave "all" the $\ce{Ca^{2+}}$ in solution that it is totally unrealistic.