Solution to Redox Reactions Problem:

equation 1: In the reactants, the O.N. of Mg is O (uncombined element), while in the products the O.N. of Mg in MgSO4 is +2 (the O.N. of ions is equal to its charge, and the ion of Mg is Mg2+)
so, the O.N. of Mg increased from O to +2, then this is oxidation. Note: O.N. stands for oxidation number.
the O.N. of H in the reactant is +1, while the O.N. of H in the products (H2 gas) is O since it's an uncombined element. the O.N. of H decreases from +1 to O, so this is reduction. Since this reactions involves a transfer of electrons between the reactants and products, then it's a redox reaction.
equation 2: In the reactants the O.N. of Cu is zero (uncombined element) while the O.N. of N in HNO3 is +5 (the sum of Oxidation number of atoms or ions in a compound is equal to zero), so O.N.(H)+O.N.(N) + 3×O.N.(O)=0 , by substituion: 1 + O.N.(N) + 3×(-2)=0 so O.N(N) = +5.
In the products, the O.N. of Cu in Cu(NO3)2 is +2 (the O.N. of ions is equal to their charge, and we have Cu2+ ion in this case), while the O.N. of nitrogen in NO is +2 (since the sum of O.N. of O and N is 0).
In sum, the oxidation number of Cu increased from O to +2, this is oxidation, while the O.N. of N decreased from +5 to +2, this is reduction. Since this reaction involves a transfer of electrons between reactants, then this is a redox reaction.
equation 3: In the reactants the O.N. of Zn is equal to O (uncombined element), while the O.N. of H is +1. In the products, the O.N. of Zn is +2 while the O.N. of H is O (uncombined element).
In sum, the O.N. of Zn increased from 0 to +2, so this is oxidation, while the O.N. of H decreased from +1 to O, so this is reduction. Since this reaction involves a transfer of electrons between reactants, then it is a redox.
equation 4: In the reactants the O.N. of H is +1 while the O.N. of Mg is O (uncombined element). In the products, the O.N. of Mg is +2 while the O.N. of H is O.
In sum, the O.N. of Mg increased from O to +2, so this is oxidation, while the O.N. of H decreased from +1 to O so this is reduction. Since this reaction involves a transfer of electrons between reactants, then it's redox