16. Given that $\left\{a_{n}\right\}$ is an arithmetic sequence with a common difference of 2, $P_{n}\left(a_{n}, b_{n}\right)$ are points on the curve $C: y=3 \cdot 2^{x-1}$, and $P_{1}$ is the intersection point of curve $C$ with the $y$-axis.
(I) Find the general terms of the sequences $\left\{a_{n}\right\}$ and $\left\{b_{n}\right\}$;
(II) Let $A_{n}=a_{2} a_{3} \cdots a_{n}, B_{n}=b_{2}+b_{3}+\cdots+b_{n}$, find the set $M$ of $n$ that satisfies $A_{n}+B_{n}<895$.
(III) Let $u=\left|\overrightarrow{P_{i} P_{j}}\right|, i, j \in M$, if $u^{2 t}+2 u^{i} \leqslant 8\left|A_{n}+B_{n}-895\right|$ holds for any $n \in \mathbf{N}^{*}$, find the range of the real number $t$.
16. Solution (I) Let $x=0$, then from $y=3 \times 2^{-1}$ we get $y=\frac{3}{2}$, so $P_{1}\left(0, \frac{3}{2}\right)$, which means $a_{1}=0, b_{1}=\frac{3}{2}$. Since $\left\{a_{n}\right\}$ is an arithmetic sequence with a common difference of 2, we have $a_{n}=2 n-2, b_{n}=3 \times 2^{2 n-3}$.
(II)
$$
\begin{aligned}
& A_{n}+B_{n}=a_{2} a_{3} \cdots a_{n}+\left(b_{2}+b_{3}+\cdots+b_{n}\right) \\
= & 2 \times 4 \times \cdots \times(2 n-2)+\left(3 \times 2+3 \times 2^{3}+\cdots+3 \times 2^{2 n-3}\right) \\
= & 2^{n-1}[1 \times 2 \times \cdots \times(n-1)]+3 \times\left(2+2^{3}+\cdots+2^{2 n-3}\right) \\
= & 2^{n}\left[3 \times 4 \times \cdots \times(n-1)+2^{n-1}\right]-2
\end{aligned}
$$

When $n=5$, $A_{5}+B_{5}=894895$.
For $n \geqslant 2$ and $n \in \mathbf{N}^{*}$, $A_{n}+B_{n}$ is monotonically increasing, so $M=\{2,3,4,5\}$.
(III) Since $u^{2 t}+2 u^{\prime} \leqslant 8\left|A_{n}+B_{n}-895\right|$ holds for $n \in \mathbf{N}^{*}$, when $n=5$, $\left|A_{n}+B_{n}-895\right|$ reaches its minimum value of 1, so $\left(u^{\prime}\right)^{2}+2 u^{\prime} \leqslant 8$, which means $\left(u^{\prime}+1\right)^{2} \leqslant 9$, thus $u^{\prime} \leqslant 2$.

Clearly, $u>1$, to make $u^{t} \leqslant 2$ hold, it is sufficient that $t \leqslant \log _{u} 2$, the key is to find the minimum value of $\log _{u} 2$. Since $\left\{a_{n}\right\}$ and $\left\{b_{n}\right\}$ are both monotonically increasing positive sequences, for $i, j \in M, \left|a_{i}-a_{j}\right| \leqslant a_{5}-a_{2}=6, \left|b_{i}-b_{j}\right| \leqslant b_{5}-b_{2}=6 \times 63$,
$$
u=\left|\overrightarrow{P_{i} P_{j}}\right|=\sqrt{\left(a_{i}-a_{j}\right)^{2}+\left(b_{i}-b_{j}\right)^{2}} \leqslant \sqrt{6^{2}+(6 \times 63)^{2}}=6 \sqrt{3970}.
$$

Therefore, the range of $t$ is $t \leqslant \log _{6 \sqrt{3970}} 2$.