Problem 69

(1) The domain of the function $f(x) = \lg\left[\log_{\frac{1}{2}}\left(\frac{1}{2}x-1\right)\right]$ is set A;
The domain of $f(x)$ satisfies $\log_{\frac{1}{2}}\left(\frac{1}{2}x-1\right) > 0$,
Therefore, $0 < \frac{1}{2}x-1 < 1$,
Therefore, $2 < x < 4$,
Therefore, set A = $(2, 4)$;
Set B = $\{x | x < 1 \text{ or } x \geq 3\}$. That is, B = $(-\infty, 1) \cup [3, +\infty)$,
Therefore, $\complement_{\mathbb{R}}B = [1, 3)$,
Hence, $A \cup B = (-\infty, 1) \cup (2, +\infty)$;
$(\complement_{\mathbb{R}}B) \cap A = (2, 3)$.
(2) From (1), we have A = $(2, 4)$; B = $(-\infty, 1) \cup [3, +\infty)$,
Since $2^a \in A$,
Therefore, $2 < 2^a < 4$,
Solving gives: $1 < a < 2$,
Also, since $\log_2(2a-1) \in B$,
Therefore, $\log_2(2a-1) < 1$ or $\log_2(2a-1) \geq 3$,
Therefore, $0 < 2a-1 < 2$ or $2a-1 \geq 8$,
Solving gives $\frac{1}{2} < a < \frac{3}{2}$ or $a \geq \frac{9}{2}$,
Therefore, $1 < a < \frac{3}{2}$.
Thus, the range of the real number $a$ is $\boxed{(1, \frac{3}{2})}$.