15. There are 40 sets of CASIO cards, each set consisting of C, A, S, I, O five cards stacked in the order of C, A, S, I, O from top to bottom. Now, these 40 sets of cards are stacked together from top to bottom, and then the first card is discarded, the second card is placed at the bottom, the third card is discarded, the fourth card is placed at the bottom, ... and so on, until only one card remains.
(1) In the above process, when only 88 cards are left, how many S cards have been discarded?
(2) What is the last remaining card, and which set does it belong to?
Three, 15. (1) 40 sets of CASIO cards total 200 cards. Number the 200 cards from top to bottom as $1,2, \cdots, 200$.

According to the operation rules, when 100 cards are discarded, the remaining card numbers are $2,4, \cdots, 200$. If another 12 cards are discarded, the involved cards are 24, with numbers $2,4, \cdots, 48$. The 12 discarded cards are $2,6,10,14,18,22,26,30,34,38,42,46$, among which, the discarded cards $\mathrm{S}$ are 2 (numbers 18 and 38). When 100 cards are discarded, there are 20 cards $\mathrm{S}$, so when only 88 cards remain, a total of 22 cards $\mathrm{S}$ have been discarded.
(2) If there are only $128\left(2^{7}\right)$ cards, the last card discarded is the one with the number 128.

Since $128<200<256$, when 72 cards are discarded, the involved cards total 144. Among the remaining 128 cards, the last card's number is 144, and $144=5 \times 28+4$, so the last card is the 4th card $\mathrm{I}$ of the 29th set.