$$\text{Let }A=\left\{x\in(0,\pi) -\left\{\frac{\pi}{2}\right\} :\log_{(2/\pi)}|\sin x| + \log_{(2/\pi)}|\cos x| = 2 \right\}\text{and }B=\left\{x\geq0 : \sqrt{x}(\sqrt{x}-4) - 3|\sqrt{x}-2| + 6 = 0 \right\}.\text{ Then } n(A\cup B) \text{ is equal to:}$$

We need to find $$n(A \cup B)$$ where:

$$A = \{x \in (0,\pi) \setminus \{\pi/2\} : \log_{2/\pi}|\sin x| + \log_{2/\pi}|\cos x| = 2\}$$

$$B = \{x \geq 0 : \sqrt{x}(\sqrt{x}-4) - 3|\sqrt{x}-2| + 6 = 0\}$$

Set A: $$\log_{2/\pi}|\sin x| + \log_{2/\pi}|\cos x| = 2$$

$$\log_{2/\pi}(|\sin x||\cos x|) = 2$$

$$|\sin x\cos x| = (2/\pi)^2 = 4/\pi^2$$

$$\frac{1}{2}|\sin 2x| = 4/\pi^2$$, so $$|\sin 2x| = 8/\pi^2$$.

Since $$8/\pi^2 \approx 0.811$$, and $$\sin 2x$$ achieves this value in $$(0, \pi)$$:

For $$x \in (0, \pi/2)$$: $$2x \in (0, \pi)$$, and $$\sin 2x = 8/\pi^2$$ has 2 solutions.

For $$x \in (\pi/2, \pi)$$: $$2x \in (\pi, 2\pi)$$, and $$|\sin 2x| = 8/\pi^2$$ has 2 solutions.

So $$|A| = 4$$.

Set B: Let $$t = \sqrt{x} \geq 0$$. The equation becomes:

$$t(t-4) - 3|t-2| + 6 = 0$$

$$t^2 - 4t + 6 - 3|t-2| = 0$$

Case 1: $$t \geq 2$$: $$t^2 - 4t + 6 - 3(t-2) = 0 \Rightarrow t^2 - 7t + 12 = 0 \Rightarrow (t-3)(t-4) = 0$$. So $$t = 3$$ or $$t = 4$$, giving $$x = 9$$ or $$x = 16$$.

Case 2: $$0 \leq t < 2$$: $$t^2 - 4t + 6 - 3(2-t) = 0 \Rightarrow t^2 - t = 0 \Rightarrow t(t-1) = 0$$. So $$t = 0$$ or $$t = 1$$, giving $$x = 0$$ or $$x = 1$$.

So $$B = \{0, 1, 9, 16\}$$, $$|B| = 4$$.

Sets A and B are disjoint (A contains values in $$(0, \pi)$$ which are irrational, B contains integers).

$$n(A \cup B) = |A| + |B| = 4 + 4 = 8$$.

The correct answer is Option B: 8.