382 cm tall, Xiao Wang's daily sleep time is uniformly distributed between 6.5 hours and 8 hours.
(1) If a day's sleep time is guaranteed to be 7 hours, then record 1, otherwise record 0. Convert the seven numbers recorded in a week into a binary number and then to a decimal number. Find the probability that the resulting number exceeds 32;
(2) Find the probability that there are at least three consecutive days in a week where the sleep time is guaranteed to be 7 hours;
(3) Find the probability that the arithmetic mean of the sleep times on Monday, Tuesday, and Wednesday exceeds 7 hours.
(1) Let the number recorded on the $i$-th day be $b_{i}(i=1,2, \cdots, 7)$. Clearly,
$$
P\left(b_{i}=1\right)=\frac{2}{3}, P\left(b_{i}=0\right)=\frac{1}{3} \text {. }
$$

The binary number $b_{1} b_{2} b_{3} b_{4} b_{5} b_{6} b_{7}$ converted to a decimal number exceeds 32 if and only if: $b_{1}=1$ or $b_{1}=0$, $b_{2}=1$, and at least one of $b_{3}, b_{4}, b_{5}, b_{6}, b_{7}$ is 1.
Note that,
$$
P\left(b_{1}=1\right)=\frac{2}{3}, b_{1}=0, b_{2}=1,
$$

and the probability that at least one of $b_{3}, b_{4}, b_{5}, b_{6}, b_{7}$ is 1 is
$$
\frac{1}{3} \times \frac{2}{3} \times\left(1-\frac{1}{3^{5}}\right) \text {. }
$$

Therefore, the required probability is
$$
\frac{2}{3}+\frac{1}{3} \times \frac{2}{3} \times\left(1-\frac{1}{3^{5}}\right)=\frac{1942}{2187} \text {. }
$$
(2) Let the event that there are at least three consecutive days of sleep in a week be $\xi$. The necessary condition for at least three consecutive days of sleep to be guaranteed for 7 hours is that $b_{3}, b_{5}$ are not both 0. Combining the value of $b_{4}$, we discuss six scenarios.
(i) The probability that $b_{3}, b_{4}, b_{5}$ are all 1 is $\left(\frac{2}{3}\right)^{3}=\frac{8}{27}$.
(ii) When $b_{3}=b_{4}=1, b_{5}=0$, it is required that $b_{2}=1$, the corresponding probability is $\frac{1}{3} \times\left(\frac{2}{3}\right)^{3}=\frac{8}{81}$.
(iii) When $b_{3}=0, b_{4}=b_{5}=1$, it is required that $b_{6}=1$, the corresponding probability is $\frac{1}{3} \times\left(\frac{2}{3}\right)^{3}=\frac{8}{81}$.
(iv) When $b_{3}=1, b_{4}=b_{5}=0$, it is required that $b_{1}=b_{2}=1$, the corresponding probability is $\left(\frac{1}{3}\right)^{2} \times\left(\frac{2}{3}\right)^{3}=\frac{8}{243}$.
(v) When $b_{3}=b_{4}=0, b_{5}=1$, it is required that $b_{6}=b_{7}=1$, the corresponding probability is $\left(\frac{1}{3}\right)^{2} \times\left(\frac{2}{3}\right)^{3}=\frac{8}{243}$.
(vi) When $b_{3}=b_{5}=1, b_{4}=0$, it is required that $b_{1}=b_{2}=1$ or $b_{6}=b_{7}=1$, the corresponding probability is
$\frac{1}{3} \times\left(\frac{2}{3}\right)^{2} \times\left[\left(\frac{2}{3}\right)^{2}+\left(\frac{2}{3}\right)^{2}-\left(\frac{2}{3}\right)^{4}\right]=\frac{224}{2187}$.
Thus, $P(\xi)=\frac{8}{27}+2 \times \frac{8}{81}+2 \times \frac{8}{243}+\frac{224}{2187}=\frac{1448}{2187}$.
(3) Let the sleep time on the $i$-th day be $a_{i}(i=1,2,3)$ hours. Let $A_{i}=\frac{a_{i}-6.5}{1.5}$. Clearly, $a_{i}$ being uniformly distributed between $6.5 \sim 8$ is equivalent to $A_{i}$ being uniformly distributed between $0 \sim 1$, and
$$
\begin{array}{l}
\frac{a_{1}+a_{2}+a_{3}}{3}>7 \\
\Leftrightarrow a_{1}+a_{2}+a_{3}>21 \\
\Leftrightarrow A_{1}+A_{2}+A_{3}>1 .
\end{array}
$$

Consider the unit cube in three-dimensional space with vertices
$$
\begin{array}{l}
O(0,0,0), B(1,0,0), C(1,1,0), \\
D(0,1,0), O^{\prime}(0,0,1), B^{\prime}(1,0,1), \\
C^{\prime}(1,1,1), D^{\prime}(0,1,1)
\end{array}
$$

Note that, $A_{i}$ being uniformly distributed between $0 \sim 1$ is equivalent to the point $P\left(A_{1}, A_{2}, A_{3}\right)$ being uniformly distributed within the unit cube. Therefore, the probability that $A_{1}+A_{2}+A_{3}>1$ is the volume of the part of the unit cube to the right and above the plane $x+y+z=1$.

The plane $x+y+z=1$ intersects the unit cube at $\triangle B D O^{\prime}$, and the part of the unit cube below and to the left of the plane $x+y+z=1$ is the tetrahedron $O B D O^{\prime}$, with a volume of $\frac{1}{6}$.
Thus, the required probability is $\frac{5}{6}$.