The dominant form of the phosphoric acid is HPO2−4 and its concentration is 0.639 mmol/L.
So, start with your three balcanced chemical equations
H3PO4+H2O⇌H2PO−4+H+3O, pKa1=2.12
H2PO−4+H2O⇌HPO2−4+H+3O, pKa2=7.21
HPO2−4+H2O⇌PO3−4+H+3O, pKa3=12.7
Because the pH of blood lies between pKa2 and pKa3, the dominant form of the acid will be HPO2−4. Because pKa2 and pKa3 vary by more than 4 units, the concentration of PO3−4 will be negligible.
So, you know that the total phosphate concentration is 1.05 mmol/L. This means that
[H2PO−4]+[HPO2−4]=1.05mmol/L (assuming [PO3−4] is negligible). (1)
Since you're dealing with a buffer, the Henderson-Hasselbalch equation can be used
#pH = pKa_2 + log(([HPO_4^(2-)])/([H_2PO_4^(-)]))#
7.40=7.21+log([HPO2−4][H2PO−4])⇒[HPO2−4][H2PO−4]=1.55
Plug this value into (1) and you'll get
[HPO2−4]1.55+[HPO2−4]=1.05, which will result in
[HPO2−4]=0.639 mmol/L
The concentration of H2PO−4 will be [H2PO−4]=0.412 mmol/L
If you're interested, you can check the assumption that [PO3−4] is negligible; the equations will produce
[PO3−4]=0.00000320 mmol/L → for all intended purposes, this is equal to zero.
One more thing...the results match the known proportions of dyhidrogen phosphate and hydrogen phosphate in extracellular fluid (39% for the former, 61% for the latter).
