Let $N$ be a given positive integer. Consider a permutation of $1,2,3,\cdots,N$, denoted as $p_1,p_2,\cdots,p_N$. For a section $p_l, p_{l+1},\cdots, p_r$, we call it "extreme" if $p_l$ and $p_r$ are the maximum and minimum value of that section. We say a permutation $p_1,p_2,\cdots,p_N$ is "super balanced" if there isn't an "extreme" section with a length at least $3$. For example, $1,4,2,3$ is "super balanced", but $3,1,2,4$ isn't. Please answer the following questions:
1. How many "super balanced" permutations are there?
2. For each integer $M\leq N$. How many "super balanced" permutations are there such that $p_1=M$?

[i]Proposed by ltf0501[/i]
### Part 1:
1. Consider a permutation of \(1, 2, 3, \ldots, N\). We need to determine how many of these permutations are "super balanced".
2. A permutation is "super balanced" if there isn't an "extreme" section with a length of at least 3. This means that for any subarray of length 3 or more, the maximum and minimum values cannot be at the ends of the subarray.
3. To construct a "super balanced" permutation, we can place the smallest number \(1\) somewhere in the permutation. The remaining numbers must be placed such that the largest and smallest numbers alternate on each side of \(1\).
4. For example, if \(1\) is placed at the leftmost position, the permutation would look like \(1, N, 2, N-1, 3, \ldots\). This ensures that no subarray of length 3 or more has the maximum and minimum values at the ends.
5. Each way we distribute the numbers into two sides, there is only one valid permutation. Therefore, the number of "super balanced" permutations is \(2^{N-1}\).

\[
\boxed{2^{N-1}}
\]

### Part 2:
1. We need to determine how many "super balanced" permutations there are such that \(p_1 = M\) for each integer \(M \leq N\).
2. If \(M = 1\) or \(M = N\), the permutation is straightforward. For \(M = 1\), the permutation is \(1, N, 2, N-1, 3, \ldots\). For \(M = N\), the permutation is \(N, 1, N-1, 2, N-2, \ldots\). In both cases, there is only one valid permutation.
3. For \(1 < M < N\), we need to place \(M\) at the first position and then distribute the remaining numbers such that the permutation remains "super balanced".
4. The numbers less than \(M\) must be placed in increasing order, and the numbers greater than \(M\) must be placed in decreasing order. This ensures that no subarray of length 3 or more has the maximum and minimum values at the ends.
5. The number of ways to choose the positions for the numbers less than \(M\) is \(\binom{N-1}{M-1}\), as we are choosing \(M-1\) positions out of \(N-1\) for the numbers less than \(M\).

\[
\boxed{\binom{N-1}{M-1}}
\]