Determine the concentration of hydrogen cations to prevent precipitation of monocalcium phosphate?
I was looking for something else and stumbled across this old problem. I noticed Martin's comment about the source, and I looked for one. Noticing that this was sponsored by ACS made me wonder if my initial analysis had been correct.

The answer key, the OP, and I agree that the $\ce{H2PO4^-}$ concentration must be $8.2\times10^{-2}$ molar to cause $\ce{Ca(H2PO4)_2}$ to start to precipitate.

Problem 1 - The problem statement lists $K_{\alpha1} = 7.1\times 10^{-3}$ but the answer key uses $7.3\times10^{-3}$

My faith in the infallibility of the ACS is shaken...

Problem 2 - There are really two ways to solve the problem. 

A. Neutralizing the acid 

The solution starts off as 0.25 molar $\ce{H3PO4}$ and 0.15 molar $\ce{Ca^{2+}}$. You add $\ce{NaOH}$ until $\ce{Ca(H2PO4)_2}$ starts to precipitate. What is the pH?

Mass balance for all phosphate species is:

$0.25 = \ce{[H3PO4] + [H2PO4^-] + [HPO4^{2-}] + [PO4^{3-}]}$

The solution is very acidic so we can assume $0 = \ce{[HPO4^{2-}] = [PO4^{3-}]}$ so

$0.25 = \ce{[H3PO4] + [H2PO4^-]}$

$\ce{[H3PO4] = 0.25 - [H2PO4^-] = 0.25 - 0.082 = 0.168 }$ 

Now using the equilibrium expression for the first ionization of phosphoric acid

$K_{\alpha1} = \dfrac{\ce{[H^+][H2PO3^-]}}{\ce{[H3PO4]}}$

Rearranging and using $7.3\times10^{-3}$ from the answer key

$\ce{[H^+]} = \dfrac{K_{\alpha1}\ce{[H3PO4]}}{\ce{[H2PO4^-]}} = \dfrac{(7.3\times10^{-3})(0.168)}{0.082} = 0.014956 \ce{-&gt;[Rounding]} 1.5\times10^{-2}$

Fortuitously using 7.1 from the problem statement yields the same rounded value.

$\dfrac{(7.1\times10^{-3})(0.168)}{0.082} = 0.014546 \ce{-&gt;[Rounding]} 1.5\times10^{-2}$

B. Making a buffer 

To the 0.25 molar solution of $\ce{H3PO4}$ you add $\ce{NaH2PO4}$ until $\ce{Ca(H2PO4)_2}$ starts to precipitate. What is the pH?

Let $x$ be the nominal molarity of the $\ce{NaH2PO4}$. 

Again the solution is very acidic so we can assume $0 = \ce{[HPO4^{2-}] = [PO4^{3-}]}$ so

$0.25 +x = \ce{[H3PO4] + [H2PO4^-]}$

But $\ce{[H2PO4^-] = 0.082}$, so 

$\ce{[H3PO4]} = 0.25 + x - \ce{[H2PO4^-]} = 0.25 + x - 0.082 = 0.168 + x$

The charge balance is given by

$\ce{[Na^+] + [H^+] = [H2PO4^-] + 2[HPO4^{2-}] + 3[HPO4^{3-}] + [OH^-]}$

Since the solution is very acidic we can neglect $\ce{[HPO4^{2-}]}$, $\ce{[HPO4^{3-}]}$, and $\ce{[OH^-]}$. So

$\ce{[Na^+] + [H^+] = [H2PO4^-]}$

$\ce{[H^+] = [H2PO4^-] - [Na^+]} = 0.082 - x$

Now substituting into the equilibrium expression for the first ionization

$K_{\alpha1} = \dfrac{\ce{[H^+][H2PO4^-]}}{\ce{[H3PO4]}} = \dfrac{(0.082-x)(0.082)}{0.168+x}$

$0.0073(0.168+x) = 0.082^2 - 0.082x$

$(0.0073 + 0.082)x = 0.0893x = 0.082^2 - 0.0071(0.168)$

$x= 0.06194$

So the buffer solution would be 0.25 molar $\ce{H3PO4}$ and 0.062 molar $\ce{NaH2PO4}$.

$\ce{[H^+] = 0.082 - 0.6194 = 0.02006 -&gt;[Rounding] 2.0\times 10^{-2}}$