Phosphate buffer system
Things to consider:


Sodium is merely a spectator ion. It will not take part in any reactions, so the initial amount of sodium you weigh in will remain constant. We can thus ignore it.
The concentrations of $\ce{H3PO4}$ and $\ce{PO4^3-}$ will be sufficiently small to entirely ignore them.


This reduces your system to the following relevant chemical equation:

$$\ce{H2PO4- + H2O &lt;=&gt; HPO4^2- + H3O+}$$

With the unknown concentrations of $\ce{H2PO4-}$, $\ce{HPO4^2-}$ and $\ce{H3O+}$. For these three we only need three equations and luckily we have them. Remember that we know:


the final phosphate concentration we want;
the final $\mathrm{pH}$ we want; and
the $\mathrm{p}K_\mathrm{a}$ value of our acidic species.


This gives us three equations that I am going to hide behind a spoiler tag:


   $[\ce{H2PO4-}] + [\ce{HPO4^2-}] = 0.1~\mathrm{M}$
 $10^{-7.198} = K_\mathrm{a} = \frac{[\ce{H3O+}][\ce{HPO4^2-}]}{[\ce{H2PO4-}]}$
 $10^{-7.4} = [\ce{H3O+}]$


We can now put the third equation into the second resulting in exactly the Henderson-Hasselbalch equation, because this exactly is how that equation was derived. The maths of one possible way to approach this problem is shown in the following spoiler tag.


   $10^{-7.198} = \frac{10^{-7.4} \times [\ce{HPO4^2-}]}{[\ce{H2PO4-}]}$
 $10^{-7.198 + 7.4} = \frac{[{HPO4^2-}]}{[\ce{H2PO4-}]}$
 $[\ce{H2PO4-}] \times 10^{0.202} = [\ce{HPO4^2-}]$ ← Here! You have the Henderson-Hasselbalch right here! (In a different form but equivalent.)

 $[\ce{HPO4^2-}] + [\ce{H2PO4-}] = 0.1~\mathrm{M}$
 $[\ce{HPO4^2-}] = 0.1~\mathrm{M} - [\ce{H2PO4-}]$

 $[\ce{H2PO4-}] \times 10^{0.202} = 0.1~\mathrm{M} - [\ce{H2PO4-}]$
 $[\ce{H2PO4-}] \left (10^{0.202} + 1 \right ) = 0.1~\mathrm{M}$
 $[\ce{H2PO4-}] = \frac{0.1~\mathrm{M}}{10^{0.202} + 1} = 0.0385~\mathrm{M}$


The last steps to calculate the required concentration of $\ce{Na2HPO4}$ including calculating the respective masses of the water-free salts should be trivial and can be found in another castle answer.

I can only speculate why you are not allowed to directly use the Henderson-Hasselbalch equation since mathematically, every optimal approach will simplify itself to exactly that equation. Maybe the point of the excersize is to derive that equation; only your prof will know.