How much NaOH required to change the pH of a weak acid solution by 2
RANT - I hate problems that don't use significant figures properly. So is the answer supposed to be +/- 1 ml, +/- 0.1 ml, or +/- 0.01 ml?
First your overall approach is right. There are three parts to solving the problem:

(1) Calculate pHs
(2) Calculate how many moles NaOH
(3) Calculate how many ml of 0.1 molar solution.

So the problem is that there are four phosphate species, $\ce{H3PO4, H2PO4^-, HPO4^{2-},}$  and $\ce{PO4^{3-}}$. The precision needed for the answer is somewhat of a mystery. So what simplifying assumptions are valid?
I'm just going to guess that the volume of NaOH should be accurate to 0.01 ml.
So let's look at the problem another way. I start off with 20.00 ml of 0.1000 molar solution of phosphoric acid and titrate with 0.01 ml at a time with a solution of 0.1000 molar sodium hydroxide. As every aliquot of NaOH is added the pH is measured. So:

(1) when I have added exactly 10.00 ml I have effectively made sodium dihydrogenphosphate. I add 2 to this pH to get the target pH.
(2) I look to see volume of NaOH that gives the target pH and subtract 10.00 ml.

SIDEBAR

Just looking at the problem I'm guessing a few ml. Let's say 5.00 ml. Measuring the NaOH to 0.01 would then mean 1 part in 500 precision (2 parts per thousand). Let's say that the initial $[\ce{H+}]$ is $3.47\times 10^{-5}$. 1 part in 347 is about right. But a pH of 4.46 only has two significant figures, that is 1 part in 46 which is too little precision for my guess of about 2 parts per thousand being needed. So the pH should be noted as 4.460.

Looking at Wikipedia we can see that for phosphoric acid that pKa1 = 2.148,
pKa2 = 7.198, and pKa3 = 12.319. 
(1) Given these pKa's we can't do better than 2 ppt. 
(2) Only a small part of the $\ce{H2PO4^-}$ will ionize to $\ce{HPO4^{2-}}$ and $\ce{H^+}$. Thus the initial pH will be lower than 7.198 but quite above 2.148.
(3) Given (2) enough $\ce{H3PO4}$ might form to lower the pH when calculating to 2 ppt. This is right on the edge for 2 ppt precision...
(4) The species $\ce{PO4^{3-}}$ can safely be ignored in all the calculations. i.e.  $\ce{[PO4^{3-}] &lt;&lt; 0.1000 molar}$
1.0 Quick and dirty method...
To avoid solving a 3rd or 4th order polynomial...
1.1 Calculate pHs
The reaction of interest is:
$\ce{H2PO4- &lt;=&gt; H+ + HPO4^{2-}}\quad\quad 6.339 \times 10^{-8}$
We'll let $\ce{[H+] = [HPO4^{2-}]}$ and assume that $\ce{[H2PO4-]} = 0.1000$ so
$ 6.339 \times 10^{-8} = \dfrac{\ce{[H+]^2}}{0.1000}$
$ \ce{[H+]} = 7.96 \times 10^{-5}$
pH = 4.099
Therefore end pH = 4.099 + 2.000 = 6.099

Checks
$\ce{[HPO4^{2−}]}$ of $7.96\times10^{−5}$ is just less than $1\times10^{-4}$ so the assumption that $\ce{[HPO4^{2−}]}$ is not appreciable to  $\ce{[H2PO4−]} \approx 0.1000$ is just ok.
We've also assumed that an insignificant amount of $\ce{H3PO4}$ will form which isn't true...  $\dfrac{\ce{[H3PO4]}}{\ce{[H2PO4^−]}} = \dfrac{\ce{[H^+]}}{7.11\times 10^{-3}} = 0.0112$

Da fix
We have 3 species of phosphate:
0.0008 $\times \ce{[H2PO4^{−}]}$ = $\ce{[HPO4^{2−}]}$
1.0000 $\times \ce{[H2PO4^{−}]}$ = $\ce{[H2PO4^{−}]}$
0.0112 $\times \ce{[H2PO4^{−}]}$ = $\ce{[H3PO4]}$
$\text{--------}$
1.0120 total $\ce{H2PO4^{−}}$
Normalizing $\ce{H2PO4^{−}}$ to 1.0000 total we have for the concentrations
0.0001 = $[\ce{HPO4^{2−}}]$
0.0988 = $[\ce{H2PO4^{−}}]$
0.001107 = $[\ce{H3PO4}]$
We rearrange the equilibrium for the reaction
$\ce{H3PO4 &lt;=&gt; H^+ + HPO4^{2−}}$
to solve for $\ce{[H^+]}$ 
$\ce{[H^+]} = \dfrac{\text{K}_{a1}\ce{[H3PO4]}}{\ce{[H2PO4^-]}} = 7.97 \times 10^{-5}$
so everything now checks...
1.2 Calculate how many moles NaOH
Given:
(1) $\ce{[H^+]} = 7.96\times 10^{-7}$
(2) 0.1 molar = $\ce{[H3PO4] + [H2PO4^-] + [HPO4^{2-}] + [PO4^{3-}]}$
Assume:
(1) $\ce{[H3PO4]} \approx 0$
(2) $\ce{[PO4^{3-}]} \approx 0$
(3) moles change in $\ce{H^+}(\text{aq})$ is insignificant
$6.339\times10^{−8} = \dfrac{\ce{[H+][HPO4^{2-}]}}{\ce{[H2PO4^-]}}$
or
$\ce{[HPO4^{2-}] =} \dfrac{6.339\times10^{−8}\ce{[H2PO4^-]}}{\ce{[H^+]}} = 0.07962\ce{[H2PO4^-]}$
We also know that
$0.1000 = \ce{[H2PO4^-] + [HPO4^{2-}] = 1.07962[H2PO4^-]}$
$\ce{[H2PO4^-] =0.09263}$
$\ce{[HPO4^{2-}] =0.00737}$
$\ce{H2PO4^- + OH^- -&gt;  HPO4^{2-}}$ so
Moles NaOH = $(0.020 \text{ liters})[0.00737] = 1.47\times10^{-4}$
1.3 Calculate how many ml of 0.1 molar solution.
liters 0.1 molar NaOH = $\dfrac{1.47\times10^{-4}}{0.1000} = 1.47\times10^{-3}$
ml 0.1 molar NaOH = 1.47