pH throughout a titration
The molarity of the $\ce{HCl}$ and the $\ce{Na2HPO4}$ is the same. So 50 ml of $\ce{HCl}$ converts the $\ce{Na2HPO4}$ to $\ce{NaH2PO4}$. The remaining 25 ml of $\ce{HCl}$ essentially converts half the $\ce{NaH2PO4}$ to $\ce{H3PO4}$. So the solution is as if 6.875 millimoles of $\ce{NaH2PO4}$ were mixed with 6.875 millimoles of $\ce{H3PO4}$ in a total of 100 ml of solution.

The Henderson-Hasselbalch equation is:

$$\text{pH} = \text{pKa} + \log{\dfrac{\ce{[H2PO4^-]}}{\ce{[H3PO4]}}} = 2.1192 + \log{\dfrac{0.06875}{0.06875}} = 2.1192 \ce{ -&gt;[Rounding] } 2.12$$

Let's check....

For the phosphate species let $C_\mathrm{P}$ be the concentration of all the phosphate species. The following equations give the fraction of each species as a function of $\ce{[H^+]}$. The equations were most easily calculated using a spreadsheet. 

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

$$\dfrac{\ce{[H3PO4]}}{[C_\mathrm{P}]} = \dfrac{\ce{[H^+]^3}}{\ce{[H^+]^3} + K_\mathrm{a1}\ce{[H^+]^2} + K_\mathrm{a1}K_\mathrm{a2}\ce{[H^+]} + K_\mathrm{a1}K_\mathrm{a2}K_\mathrm{a3}} = 0.49999$$

$$\dfrac{\ce{[H2PO4^-]}}{[C_\mathrm{P}]} = \dfrac{\ce{K_\mathrm{a1}[H^+]^2}}{\ce{[H^+]^3} + K_\mathrm{a1}\ce{[H^+]^2} + K_\mathrm{a1}K_\mathrm{a2}\ce{[H^+]} + K_\mathrm{a1}K_\mathrm{a2}K_\mathrm{a3}} = 0.50001 $$

$$\dfrac{\ce{[HPO4^{2-}]}}{[C_\mathrm{P}]} = \dfrac{\ce{K_\mathrm{a1}K_\mathrm{a2}[H^+]}}{\ce{[H^+]^3} + K_\mathrm{a1}\ce{[H^+]^2} + K_\mathrm{a1}K_\mathrm{a2}\ce{[H^+]} + K_\mathrm{a1}K_\mathrm{a2}K_\mathrm{a3}} = 4.0791\times10^{-06}$$

$$\dfrac{\ce{[PO4^{3-}]}}{[C_\mathrm{P}]} = \dfrac{K_\mathrm{a1}K_\mathrm{a2}K_\mathrm{a3}}{\ce{[H^+]^3} + K_\mathrm{a1}\ce{[H^+]^2} + K_\mathrm{a1}K_\mathrm{a2}\ce{[H^+]} + K_\mathrm{a1}K_\mathrm{a2}K_\mathrm{a3}} = 1.1272\times10^{-16} $$

Thus the assumption that the H_H equation can be used is valid. 

The textbook answer is wrong.