(a) Let's list the first 20092009 natural numbers. Then, we successively replace each number with the sum of its digits, until we obtain a list of numbers with only one digit. Does the list have more digits 4 or 5? How many 9s are there in the list?

(b) Applying the same process to the number $3^{2009}$, that is, replacing the number with the sum of its digits, what is the single-digit number obtained?

(c) And for the number $17^{2009}$?
(a) We know that a number and the sum of its digits always leave the same remainder when divided by 9. Thus, the difference between the initial number and the final number is always divisible by 9.

By performing the steps successively, we obtain the digits from 1 to 9. Therefore, the final list is:

$$
1,2,3,4,5,6,7,8,9,1,2,3, \ldots
$$

Since the remainder of the division of the number 20092009 by 9 is 4, the last 6 digits of the list are: ..., $8,9,1,2,3,4$. Therefore, the list has more 4s than 5s.

The number of times the digit 9 appears in the list is the number of multiples of 9 that are less than or equal to 20092009. Since 20092005 is the largest multiple of 9 that is less than 20092009, we have that $20092005 \div 9=2232445$ times the digit 9 appears in the list.

(b) Since $3^{2009}=3^{2008} \times 3=\left(3^{2}\right)^{1004} \times 3=9^{1004} \times 3$, the remainder of the division by 9 is 0. Therefore, the final single-digit number is 9.

(c) Observe that $17^{2}=$ multiple of $9+1$. Thus,

$$
17^{2008}=\left(17^{2}\right)^{1004}=\text { multiple of } 9+1
$$

so $17^{2009}=$ multiple of $9+17=$ multiple of $9+8$.

Therefore, we can conclude that if we perform the same process with the number $17^{2009}$, we obtain the digit 8 at the end.