Here's my reasoning:

The 1st ionisation is:

#sf(H_3A^(2+)rightleftharpoonsH_2A^(+)+H^+)#

For which:

#sf(K_(a1)=([H_2A^(+)][H^(+)])/([H_3A^(2+)]))#

You can see that as the pH rises when alkali is added there comes a point when #sf([H_2A^(+)]=[H_3A^(2+)])#. The expression becomes:

#sf(K_(a1)=(cancel([H_2A^(+)])[H^(+)])/(cancel([H_3A^(2+)])))#

So

#sf(K_(a1)=[H^(+)])#

And

#sf(pK_(a1)=pH)#

This tells us that at pH 3.5 the acid is 1/2 neutralised and we will have an equlimolar mixture of #sf(H_3A^(2+))# and #sf(H_2A^(+))#.

At the 1st equivalence point the solution will contain #sf(H_2A^(+))#

The 2nd ionisation is:

#sf(H_2A^(+)rightleftharpoonsHA+H^(+))#

By the same reasoning when #sf(pK_(a2)=pH=6.0)# there will be an equilmolar mixture of #sf(H_2A^(+))# and #sf(HA)#.

We would expect the proportion of #sf(H_2A^(+))# to #sf(HA)# to decrease as the pH is raised.

The expression for #sf(K_(a2))# is:

#sf(K_(a2)=([HA][H^(+)])/([H_2A^(+)]))#

Rearranging:

#sf([H^+]=K_(a2)xx([H_2A^(+)])/([HA]))#

Taking -ve logs of both sides gives:

#sf(pH=pK_(a2)-log([[H_2A^(+)]]/([HA]))#

#:.##sf(7.4=6.0-log([[H_2A^(+)]]/[[HA]))#

#:.##sf(log([[H_2A^(+)]]/[[HA]])=6.0-7.4=-1.4)#

From which:

#sf(([H_2A^(+)])/([HA])=0.0398)#

This means the solution is about 4% #sf(H_2A^(+))#.

I'm not sure what is meant by "the average charge" so suggest it would be about 0.04.