(Example 2 Given the inequality about $x$: $k \cdot 4^{x}-2^{x+1}+6 k<0$.
(1) If the solution set of the inequality is $\left\{x \mid 1<x<\log _{2} 3\right\}$, find the value of the real number $k$;
(2) If the solution set of the inequality is a subset of $\left\{x \mid 1<x<\log _{2} 3\right\}$, find the range of the real number $k$;
(3) If the inequality holds for all $x \in\left\{x \mid 1<x<\log _{2} 3\right\}$, find the range of the real number $k$.
Let $t=2^{x}$, then the original inequality becomes $k t^{2}-2 t+6 k<0$, and
$$
\left\{\begin{array}{l}
\Delta=4-24 k^{2}>0 \\
2+3=\frac{2}{k} \\
2 \cdot 3=\frac{6 k}{k}
\end{array}\right.
$$

We get $k=\frac{2}{5}$.
(2) The solution set of the original inequality is $\left\{x \mid 10$ when, if $\Delta=4-24 k^{2} \leqslant 0$, i.e., $k^{2} \geqslant \frac{1}{6}, k \geqslant \frac{\sqrt{6}}{6}$, then the solution set of the inequality is $\varnothing$, $\varnothing \subseteq(2,3)$ holds; if $\Delta>0$, let $f(t)=k t^{2}-2 t+6 k$, then
$$
\left\{\begin{array}{l}
\Delta=4-24 k^{2}>0, \\
f(2)=4 k-4+6 k \geqslant 0, \\
f(3)=9 k-6+6 k \geqslant 0, \\
20$ is consistent with the problem;
When $k0$, let $g(t)=t^{2}-\frac{2}{k} t+6$, the axis of symmetry is $t=\frac{1}{k}0$, then for all $20$ it holds.
When $k>0$, the inequality is $t^{2}-\frac{2}{k} t+6<0$, let $g(t)=t^{2}-\frac{2}{k} t+6$, the axis of symmetry
In summary, $k \leqslant \frac{2}{5}$.
Key points for solving the problem: First, use substitution to transform the exponential equation into a simple quadratic equation with a letter coefficient. The first part (1) is the inverse problem of solving inequalities, directly applying the relevant tools of quadratic equations; The second and third parts (2) and (3) involve the combination of quadratic functions and graphs, examining the problem intuitively and vividly. The results can vary greatly, so careful reading of the problem is crucial.