10ml of .1M Diprotic Acid titrated with .1M NaOH pka1=1.87 pka2=6.07

Looks like maleic

Vol. req. to reach eq pt 1 = 10ml

[HA-]=(.1M x .010L)/ 0.02L = 0.05M

So far so good.

HA- + H2O <-> H2A +OH-

pka1 = 10^(-1.87) =1.35E-2

ka = x^2 / ( 0.05-x )

6.75E-4 -1.35E-2x = x^2

x^2 + 1.35E-2x- 6.75E-4

x=0.0200

x=-0.0335

pH = -log (0.02) = 1.69

No. HA

^{-} is amphiprotic, thus it is here that you should use (pKa1+pKa2)/2

pH second equivalence point is .5(ka1 + ka2) do this first?

to solve the at the second equivalence point equatuion is this .5(ka1 + ka2) should i solve this first?

Sorry, my English fails here.

At the second equivalence point you have just A

^{2-}. This is multiprotic base, but difference in stregth of both steps of basic dissociation is large enough - you may treat it as if only first dissociation step occurs.

Calculate ppH for both equivalence points using approach you should be aware off atm. Download my BATE and play with maleic acid/sodium hydroxide solution, to see how pH changes. Compare values displayed by the program with the results of your calculations, to see when simplified approach fails.