Here is example 4b:

KMnO4 + KCl + H2SO4 ---> K2SO4 + MnSO4 + Cl2 + H2O

1) identify the atoms being oxidized and reduced:

Mn goes from +7 to +2; it is reduced
Cl goes from -1 to zero; it is oxidized

2) write the two half-reactions:

Mn7+ + 5e¯ ---> Mn2+
2Cl¯ ---> Cl2 + 2e¯

3) multiply the first reaction by 2 and the second by 5, and add them to get:

2Mn7+ + 10Cl¯ ---> 2Mn2+ + 5Cl2

4) since this a complete redox reaction and the number of electrons are equal on both sides, we can now use these coefficients in the molecular equation to balance it:

2KMnO4 + 10KCl + H2SO4 ---> K2SO4 + 2MnSO4 + 5Cl2 + H2O
5) balancing the other atoms can be done easily now by 'trial and error' method:
This is left to the reader.