(a) To divide the remaining 12 members into cars of 6, 5, and 4 people, we calculate:

\(\binom{12}{6} \times \binom{6}{5} \times \binom{1}{1} = 924 \times 6 \times 1 = 5544\)

However, considering the arrangement of cars, the correct calculation is:

\(\frac{12!}{6! \times 5! \times 4!} = 27720\)

(c) To choose a team of 4 adults and 3 children from different families:

Choose 4 adults from 7: \(\binom{7}{4} = 35\)

Choose 3 children from different families: \(3! = 6\)

Total ways: \(35 \times 6 = 210\)

However, the correct calculation is:

\(\binom{7}{4} \times \binom{3}{1} \times \binom{3}{1} \times \binom{3}{1} = 420\)

(d) To ensure at least one of Mr Kenny or Mr Lizo is included:

Total ways without restriction: \(\binom{7}{4} \times \binom{8}{3} = 1960\)

Ways excluding both Mr Kenny and Mr Lizo: \(\binom{5}{4} \times \binom{8}{3} = 280\)

Thus, ways including at least one: \(1960 - 280 = 1680\)