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 level is considered passed. The questions are:
(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 the numbers $1,2,3,4,5,6$ on its faces. The number of points on the face that lands up after rolling the die is the number of points that appear.)
Three, 13. Since the die is a uniform cube, the possibility of each point number appearing after throwing is the same.
(1) Because the maximum point number that the die can show is 6, and $6 \times 4 > 2^{4}$, $6 \times 5 < 2^{5}$, therefore, when $n \geqslant 5$, the sum of the point numbers appearing $n$ times being greater than $2^{n}$ is impossible. Hence, this is an impossible event, and the probability of ultimately passing the level is 0. Therefore, the maximum number of consecutive levels that can be passed is 4.
(2) Let event $A_{n}$ be "failing the $n$-th level," then the complementary event $\overline{A_{n}}$ is "successfully passing the $n$-th level."
In the $n$-th level game, the total number of basic events is $6^{n}$.
Level 1: The number of basic events contained in event $A_{1}$ is 2 (i.e., the point number is 1 or 2 in these two cases). Therefore, the probability of passing this level is
$$
P\left(\overline{A_{1}}\right)=1-P\left(A_{1}\right)=1-\frac{2}{6}=\frac{2}{3} \text{. }
$$

Level 2: The number of basic events contained in event $A_{2}$ is the sum of the positive integer solutions of the equation $x+y=a$ when $a$ takes the values 2, 3, 4, which is $\mathrm{C}_{1}^{1}+\mathrm{C}_{2}^{1}+\mathrm{C}_{3}^{1}=6$ (cases). Therefore, the probability of passing this level is
$$
P\left(\overline{A_{2}}\right)=1-P\left(A_{2}\right)=1-\frac{6}{6^{2}}=\frac{5}{6} \text{. }
$$

Level 3: The basic events contained in event $A_{3}$ are the sum of the positive integer solutions of the equation $x+y+z=a$ when $a$ takes the values 3, 4, 5, 6, 7, 8, which is $C_{2}^{2}+C_{3}^{2}+C_{4}^{2}+C_{5}^{2}+C_{6}^{2}+C_{7}^{2}=56$ (cases). Therefore, the probability of passing this level is
$$
P\left(\overline{A_{3}}\right)=1-P\left(A_{3}\right)=1-\frac{56}{6^{3}}=\frac{20}{27} \text{. }
$$

Thus, the probability of passing the first three levels consecutively is
$$
P\left(\overline{A_{1}}\right) \times P\left(\overline{A_{2}}\right) \times P\left(\overline{A_{3}}\right)=\frac{2}{3} \times \frac{5}{6} \times \frac{20}{27}=\frac{100}{243} \text{. }
$$

Note: The number of basic events in the 2nd and 3rd levels can also be listed.