solution problem 1 : NOTE: that the compounds in both (a) and (b) contain metal and non metal atoms, so we expect them to be ionic compounds.The first thing to do is write the formulas of the cation and anion, then name the ions .
(a) Mg group IIA is expected to form only a +2 ion Mg2+ , the magnesium ion . Nitrogen ( group VA) is expected to form an anion of charge equal to the group number minus 8 so N3- ion .Name of the cation followed by anion , name of Mg3N2 is magnesium nitride.
(b) Chromium is the transition element and like the most such elements, has more than one monoatomic ion. You can find the charge on the Cr ion if you know the formula of anion.SO4 in the CrSO4 from the table of polyatomic ion is SO42- ion .Therefore Cr2+ must be the cation to give electrical neutrality to the compound. Cr2+ IS THE chromium (II) ion , so the name of compound is chromium (II) sulfate.
Solution to problem 2 :
(a) Iron (II) phosphate : contains the iron (II) ion , Fe2+ and phosphate ion PO43- .Now you know the formula of ions, use the same method ( you magnitude of charge on one ion to obtain the subscript for other ion). The formula is Fe3(PO4)2.
(b) Titanium (IV) oxide is composed of titanium (IV) ions, Ti4+ and oxide ions O2- .The formula is titanium (IV) oxide.