Example 6 Given the function $f(x)=x+\frac{t}{x}(t>0)$ and the point $P(1,0)$, draw two tangent lines $P M$ and $P N$ from $P$ to the curve $y=f(x)$, with the points of tangency being $M$ and $N$.
(1) Let $|M N|=g(t)$, find the expression for the function $g(t)$.
(2) Does there exist a $t$ such that $M$, $N$, and $A(0,1)$ are collinear? If so, find the value of $t$; if not, explain the reason.
(3) Under the condition in (1), if for any positive integer $n$, there exist $m+1$ real numbers $a_{1}, a_{2}, \cdots, a_{m}, a_{m+1}$ in the interval $\left[2, n+\frac{64}{n}\right]$ such that the inequality
$$
g\left(a_{1}\right)+g\left(a_{2}\right)+\cdots+g\left(a_{m}\right)<g\left(a_{m+1}\right)
$$
holds, find the maximum value of $m$.
Solution: (1) The equation of the chord of tangents $MN$ passing through the external point $P(1,0)$ of the curve $y=x+\frac{t}{x}(t>0)$ is
$$
y=2 x+2 t \text {. }
$$

Substituting into $y=x+\frac{t}{x}(t>0)$, we get $x^{2}+2 t x-t=0$.
Thus, $x_{1}+x_{2}=-2 t, x_{1} x_{2}=-t \ldots$
Therefore, $g(t)=\sqrt{\left(x_{1}-x_{2}\right)^{2}+\left(x_{1}+\frac{t}{x_{1}}-x_{2}-\frac{t}{x_{2}}\right)^{2}}$
$$
=\sqrt{5}\left|x_{1}-x_{2}\right|=2 \sqrt{5 t^{2}+5 t}(t>0) \text {. }
$$
(2) When $M, N$ and $A(0,1)$ are collinear, substituting $A(0,1)$ into $y=2 x+2 t$, we get $t=\frac{1}{2}$.
(3) It is easy to see that $g(t)$ is an increasing function on the interval $\left[2, n+\frac{64}{n}\right]$, then
$$
\begin{array}{l}
m g(2) \leqslant g\left(a_{1}\right)+g\left(a_{2}\right)+\cdots+g\left(a_{m}\right) \\
<g\left(a_{m+1}\right) \leqslant g\left(n+\frac{64}{n}\right)
\end{array}
$$

holds for all positive integers $n$, i.e.,
$$
m<\sqrt{\frac{1}{6}\left[\left(n+\frac{64}{n}\right)^{2}+\left(n+\frac{64}{n}\right)\right]}
$$

holds for all positive integers $n$.
Since $n+\frac{64}{n} \geqslant 16$, we have
$$
\sqrt{\frac{1}{6}\left[\left(n+\frac{64}{n}\right)^{2}+\left(n+\frac{64}{n}\right)\right]} \geqslant \sqrt{45 \frac{1}{3}} \text {. }
$$

Since $m$ is a positive integer, we take $m=\left[\sqrt{45 \frac{1}{3}}\right]=6$.
When $m=6$, there exist
$$
a_{1}=a_{2}=\cdots=a_{m}=2, a_{m+1}=16 \text {, }
$$

satisfying the condition for all $n$, thus, the maximum value of $m$ is 6.