8. C.

Obviously, if Ricardo has the most 5 points, then he has the largest amount of money, which is $2019 \times 5+1$ points.

Similarly, if he has the most 1 point, then he has the smallest amount of money, which is $2019 \times 1+5$ points.

Therefore, the difference between the maximum and minimum amount of money he can have is
$$
\begin{array}{l}
(2019 \times 5+1)-(2019 \times 1+5) \\
=2019 \times 4-4=2018 \times 4=8072 .
\end{array}
$$
8. Ricardo has 2020 coins, some of which are 1 cent, and the rest are 5 cents, and there is at least one 1 cent coin and at least one 5 cent coin. Then the difference between the maximum and minimum possible amounts of money Ricardo can have is ( ) cents.
(A) 8062
(B) 8068
(C) 8072
(D) 8076
(E) 8082