Solution:

We will denote the people by $A, B, C, \ldots$ and their initial balls by the corresponding small letters. Thus the initial state is $A a, B b, C c, D d, E e(, F f)$. A swap is denoted by the (capital) letters of the people involved.

a) Five people form 10 pairs, so at least 10 swaps are necessary.

In fact, 10 swaps are sufficient:

Swap $A B$, then $B C$, then $C A$; the state is now $A a, B c, C b, D d, E e$.

Swap $A D$, then $D E$, then $E A$; the state is now $A a, B c, C b, D e, E d$.

Swap $B E$, then $C D$; the state is now $A a, B d, C e, D b, E c$.

Swap $B D$, then $C E$; the state is now $A a, B b, C c, D d, E e$.

All requirements are fulfilled now, so the answer is 10.

b) Six people form 15 pairs, so at least 15 swaps are necessary. We will prove that the final number of swaps must be even. Call a pair formed by a ball and a person inverted if letter of the ball lies after letter of the person in the alphabet. Let $T$ be the number of inverted pairs; at the start we have $T=0$. Each swap changes $T$ by 1, so it changes the parity of $T$. Since in the end $T=0$, the total number of swaps must be even. Hence, at least 16 swaps are necessary. In fact 16 swaps are sufficient:

Swap $A B$, then $B C$, then $C A$; the state is now $A a, B c, C b, D d, E e, F f$.

Swap $A D$, then $D E$, then $E A$; the state is now $A a, B c, C b, D e, E d, F f$.

Swap $F B$, then $B E$, then $E F$; the state is now $A a, B d, C b, D e, E c, F f$.

Swap $F C$, then $C D$, then $D F$; the state is now $A a, B d, C e, D b, E c, F f$.

Swap $B D$, then $C E$, then twice $A F$, the state is now $A a, B b, C c, D d, E e, F f$.

All requirements are fulfilled now, so the answer is 16.