(a) The problem is that there are 8 positions where 30 different things can be put, and the methods are also in sequence, so A is 8 above and 30 below.

(2) This question seems a bit ambiguous? The question is: How many ways do ab win prizes or how many ways do the top eight winners win prizes? If it is the former, it should be easy to calculate, so it is more likely to estimate the latter. Then there are three ways for ab to be in the top three, namely, the top two and the bottom three of A, which is equivalent to completing the first step of the whole problem. Ab has occupied two positions, so it is equivalent to leaving six positions, which need six of the remaining 28 people to occupy, so it is the first six and the last 28 of A, which is equivalent to completing the second step of the whole problem, so the final answer is (the first two and the last three of A) using the principle of residue method.

(2)

First question, the first symbol can have 40 choices, and the second symbol can also have 40 choices, so if a message consists of 25 symbols, then the total number * * * can be sent to the 25th power of 40.

Second, the first symbol can have 10 choices, and the second symbol can also have 10 choices. From the third symbol to the 40th symbol, each symbol has 30 choices, so send10 *10 * 30 * ... = 65448.

(3) Will there be some problems? It is said in (b) that if a letter cannot appear twice, how can it be a palindrome? And the concept of palindrome is not given in the title? Sorry, the third question is a bit ambiguous and I can't help you answer it.