Q2
Ca and Mg cations react identically with phospate, hence having molar concentrations you can add them during stoichiometry, ie
you have 0.050 + 0.095 = 0.145 M (Ca+Mg)
3Ca2+ + 2PO43- = (Ca,Mg)3(PO4)2
hence you need 2x0.145/3=0.097 mole of anhydrous phosphate (which is about 16 g - crude approximation!)
Sodium phosphate is Na3PO4
Note this is shortcut for advanced students. You need rather Borek hint