(5) The positive sequence $\left\{a_{n}\right\}$ satisfies $\frac{1}{a_{n} a_{n+1}}+\frac{1}{a_{n} a_{n+2}}+\frac{1}{a_{n+1} a_{n+2}}=1\left(n \in \mathbf{N}^{*}\right)$, $a_{1}+a_{3}=6, a_{1} 、 a_{2} 、 a_{3}$ are in increasing geometric progression, $S_{n}$ is the sum of the first $n$ terms of $\left\{a_{n}\right\}$, then the value of $\left.S_{2008}\right]$ is ( $[x]$ denotes the greatest integer not exceeding the real number $x$)( ).
(A) 5352
(B) 5353
(C) 5357
(D) 5358
(5) From the given, we have
$$
a_{n}+a_{n+1}+a_{n+2}=a_{n} a_{n+1} a_{n+2},
$$

Therefore,
$$
a_{1}+a_{2}+a_{3}=a_{1} a_{2} a_{3},
$$

And $a_{1} a_{3}=a_{2}^{2}$, so $a_{2}^{3}=a_{2}+6$, which implies
$$
\left(a_{2}-2\right)\left(a_{2}^{2}+2 a_{2}+3\right)=0,
$$

Solving this, we get $a_{2}=2$. Since $a_{1}, a_{2}, a_{3}$ form a geometric sequence, we have $a_{1}+\frac{4}{a_{1}}=6$, solving this gives $a_{1}=$ $3 \pm \sqrt{5}$. Given that $a_{1}, a_{2}, a_{3}$ are monotonically increasing, we have $a_{1}=3-\sqrt{5}$.
From (1), we get
$$
a_{n+1}+a_{n+2}+a_{n+3}=a_{n+1} a_{n+2} a_{n+3},
$$
(2) - (1) gives
$$
\left(a_{n+3}-a_{n}\right)\left(1-a_{n+1} a_{n+2}\right)=0,
$$

Thus, $a_{n}=a_{n+3}$, so $\left\{a_{n}\right\}$ is a sequence with a period of 3. Therefore, $a_{2008}=a_{1}=3-\sqrt{5}$, and thus
$$
\begin{aligned}
S_{2008} & =S_{2007}+a_{2008} \\
& =\frac{2007}{3} \times(6+2)+(3-\sqrt{5}) \\
& =5352+(3-\sqrt{5}),
\end{aligned}
$$

Hence, $\left[S_{2008}\right]=5352$. Therefore, the answer is A.