13. The rules of a "level-passing game" stipulate: on the $n$-th level, a die must be rolled $n$ times. If the sum of the points from these $n$ rolls is greater than $2^n$, the player passes the level. Questions:
(1) What is the maximum number of levels a person can pass in this game?
(2) What is the probability that he can pass the first three levels consecutively?
(Note: A die is a uniform cube with points numbered $1,2,3,4,5,6$ on its faces. The point number on the face that lands up after the die is rolled and comes to rest is the point number that appears.)
(1) Each time a die is rolled, the maximum number of points that can appear is 6, so the maximum sum of points that can appear in $n$ rolls of the die in the $n$-th level is $6n$. According to the problem, $6n > 2^n \Rightarrow n \leqslant 4$.

Therefore, a person can pass a maximum of 4 levels in this game.
(2) Let the probabilities of the person passing the 1st, 2nd, and 3rd levels be $p_{1}, p_{2}, p_{3}$, respectively. Since there are 2 outcomes where the number of points on a single roll of the die is less than or equal to 2; there are 6 outcomes where the sum of points from 2 rolls of the die is less than or equal to 4; and there are 56 outcomes where the sum of points from 3 rolls of the die is less than or equal to 8. Thus, $p_{1}=\frac{4}{6}=\frac{2}{3} ; p_{2}=\frac{30}{36}=\frac{5}{6} ; p_{3}=\frac{160}{216}=\frac{20}{27}$. Since the events of the person passing a certain level are independent, let $A=$ “he passes the first three levels consecutively”, so $P(A)=p_{1} p_{2} p_{3}=\frac{2}{3} \cdot \frac{5}{6} \cdot \frac{20}{27}=\frac{100}{243}$.