Let $ n\geq 3$ be integer. Throw a  dice $ n$ times, let $ X_1,\ X_2,\ \cdots ,\ X_n$ be the number of the outcome in this order respectively. For $ i \equal{} 2,\ 3,\ \cdots ,\ n$, denote $ A_i$ the event such that $ X_i \equal{} X_{i \minus{} 1}$.

(1) What's the probability $ p_n$ that at least one of $ A_2,\ A_3,\ \cdots ,\ A_n$ occurs?

(2) What's the probability $ q_n$ that at least two of $ A_2,\ A_3,\ \cdots ,\ A_n$ occurs?
### Part (1)
1. We need to find the probability \( p_n \) that at least one of the events \( A_2, A_3, \ldots, A_n \) occurs. 
2. First, we calculate the probability that none of these events occur. This means that all consecutive dice throws are different.
3. The probability that \( X_i \neq X_{i-1} \) for any \( i \) is \( \frac{5}{6} \).
4. Since there are \( n-1 \) such pairs, the probability that all consecutive dice throws are different is \( \left(\frac{5}{6}\right)^{n-1} \).
5. Therefore, the probability that at least one of the events \( A_2, A_3, \ldots, A_n \) occurs is:
   \[
   p_n = 1 - \left(\frac{5}{6}\right)^{n-1}
   \]

### Part (2)
1. We need to find the probability \( q_n \) that at least two of the events \( A_2, A_3, \ldots, A_n \) occur.
2. First, we calculate the probability that exactly one of these events occurs.
3. The probability that \( A_2 \) occurs and none of the others occur is:
   \[
   \frac{1}{6} \times \left(\frac{5}{6}\right)^{n-2}
   \]
4. The probability that \( A_3 \) occurs and none of the others occur is:
   \[
   \frac{5}{6} \times \frac{1}{6} \times \left(\frac{5}{6}\right)^{n-3}
   \]
5. Similarly, the probability that \( A_n \) occurs and none of the others occur is:
   \[
   \left(\frac{5}{6}\right)^{n-2} \times \frac{1}{6}
   \]
6. Summing these probabilities, the probability that exactly one of \( A_2, A_3, \ldots, A_n \) occurs is:
   \[
   (n-1) \times \frac{1}{6} \times \left(\frac{5}{6}\right)^{n-2}
   \]
7. Therefore, the probability that at least two of the events \( A_2, A_3, \ldots, A_n \) occur is:
   \[
   q_n = 1 - \left(\frac{5}{6}\right)^{n-1} - (n-1) \times \frac{1}{6} \times \left(\frac{5}{6}\right)^{n-2}
   \]
8. Simplifying the expression, we get:
   \[
   q_n = 1 - \left(\frac{5}{6}\right)^{n-1} - \frac{(n-1) \times 5^{n-2}}{6^{n-1}}
   \]
   \[
   q_n = 1 - \frac{5^{n-1}}{6^{n-1}} - \frac{(n-1) \times 5^{n-2}}{6^{n-1}}
   \]
   \[
   q_n = 1 - \frac{5^{n-2} \times (5 + n - 1)}{6^{n-1}}
   \]
   \[
   q_n = 1 - \frac{5^{n-2} \times (n + 4)}{6^{n-1}}
   \]

The final answer is \( \boxed{ p_n = 1 - \left(\frac{5}{6}\right)^{n-1} } \) and \( q_n = 1 - \frac{5^{n-2} \times (n + 4)}{6^{n-1}} \)