【Knowledge Points】Sequence recursion formula; Determination of arithmetic relationship. D1 D2
【Answer Analysis】(I) See analysis (II) $\lambda=4$.
Analysis: (I) Given, $a_na_{n+1}=\lambda S_n-1$, $a_{n+1}a_{n+2}=\lambda S_{n+1}-1$.
Subtracting the two equations, we get $a_{n+1}(a_{n+2}-a_n)=\lambda a_{n+1}$.
Since $a_{n+1} \neq 0$, it follows that $a_{n+2}-a_n=\lambda$.
(II) Given, $a_1=1$, $a_1a_2=\lambda S_1-1$, we can obtain $a_2=\lambda-1$.
From (I), we know $a_3=\lambda+1$.
Letting $2a_2=a_1+a_3$, we solve to get $\lambda=4$.
Therefore, $a_{n+2}-a_n=4$, from which we can deduce
$\{a_{2n-1}\}$ is an arithmetic sequence with the first term as 1 and common difference as 4, $a_{2n-1}=4n-3$;
$\{a_{2n}\}$ is an arithmetic sequence with the first term as 3 and common difference as 4, $a_{2n}=4n-1$.
Thus, $a_n=2n-1$, $a_{n+1}-a_n=2$.
Therefore, when $\lambda=4$, the sequence $\{a_n\}$ is an arithmetic sequence.
【Thought Process】(I) Use $a_na_{n+1}=\lambda S_n-1$ and $a_{n+1}a_{n+2}=\lambda S_{n+1}-1$, subtract to derive the result;
(II) First, from the given, we can obtain $a_2=\lambda-1$. Knowing from (I) that $a_3=\lambda+1$, we solve to get $\lambda=4$, then judge accordingly.

Thus, the final answers are:
- For part (I), the proof that $a_{n+2}-a_n=\lambda$.
- For part (II), the sequence $\{a_n\}$ is an arithmetic sequence when $\lambda=\boxed{4}$.