7. (i) (Grade 11) Let $M=\{1,2, \cdots, 100\}, A$ be a subset of $M$, and $A$ contains at least one cube number. Then the number of such subsets $A$ is $\qquad$
(ii) (Grade 12) Two players, A and B, are playing a table tennis singles final, using a best-of-five format (i.e., the first to win three games wins the championship). For each game, the probability of A winning is $\frac{2}{3}$, and the probability of B winning is $\frac{1}{3}$. Then the probability of an upset (B winning the championship) is $\qquad$ .
(ii) $2^{100}-2^{96}$ (i.e., $\left.15 \times 2^{96}\right)$.
$M$ contains 4 cube numbers, which are $1^{3}, 2^{3}, 3^{3}, 4^{3}$. The total number of subsets of $M$ is $2^{100}$, among which, the number of subsets that do not contain cube numbers is $2^{96}$. Therefore, the number of subsets that contain cube numbers is $2^{100}-2^{96}$.
(ii) $\frac{17}{81}$.
If one side wins 3:1 or 3:0, then the unplayed one or two games can be added without affecting the result. Thus, the problem is transformed into:
Finding the probability that player B wins at least 3 out of 5 games.
The probability that B wins all 5 games is $\left(\frac{1}{3}\right)^{5}$;
The probability that B wins 4 games and loses 1 game is $\mathrm{C}_{5}^{1}\left(\frac{1}{3}\right)^{4} \times \frac{2}{3}$;
The probability that B wins 3 games and loses 2 games is $\mathrm{C}_{5}^{2}\left(\frac{1}{3}\right)^{3} \times\left(\frac{2}{3}\right)^{2}$.
Adding these probabilities, the sum is $\frac{17}{81}$.