Let S = \ θ ∈ (0, (π)/(2)) : ∑_ m=1 ^ 6 (θ + (m-1)(π)/(4)) (θ +...

Let the given sum be $E$. We can write the product of cosecants as:

$E = \sum_{m=1}^{6} \dfrac{1}{\sin\left(\theta + (m-1)\dfrac{\pi}{4}\right) \sin\left(\theta + \dfrac{m\pi}{4}\right)}$

Multiplying and dividing by $\sin\left(\dfrac{\pi}{4}\right) = \dfrac{1}{\sqrt{2}}$, we get:

$E = \sqrt{2} \sum_{m=1}^{6} \dfrac{\sin\left(\left(\theta + \dfrac{m\pi}{4}\right) - \left(\theta + (m-1)\dfrac{\pi}{4}\right)\right)}{\sin\left(\theta + (m-1)\dfrac{\pi}{4}\right) \sin\left(\theta + \dfrac{m\pi}{4}\right)}$

$E = \sqrt{2} \sum_{m=1}^{6} \left[ \cot\left(\theta + (m-1)\dfrac{\pi}{4}\right) - \cot\left(\theta + \dfrac{m\pi}{4}\right) \right]$

This is a telescoping sum. Expanding the terms:

$E = \sqrt{2} \left[ \left(\cot\theta - \cot\left(\theta + \dfrac{\pi}{4}\right)\right) + \left(\cot\left(\theta + \dfrac{\pi}{4}\right) - \cot\left(\theta + \dfrac{2\pi}{4}\right)\right) + \dots + \left(\cot\left(\theta + \dfrac{5\pi}{4}\right) - \cot\left(\theta + \dfrac{6\pi}{4}\right)\right) \right]$

$E = \sqrt{2} \left[ \cot\theta - \cot\left(\theta + \dfrac{3\pi}{2}\right) \right]$

Using the allied angle identity $\cot\left(\theta + \dfrac{3\pi}{2}\right) = -\tan\theta$, we have:

$E = \sqrt{2} (\cot\theta + \tan\theta) = \sqrt{2} \left( \dfrac{\cos\theta}{\sin\theta} + \dfrac{\sin\theta}{\cos\theta} \right) = \dfrac{\sqrt{2}}{\sin\theta \cos\theta} = \dfrac{2\sqrt{2}}{\sin(2\theta)}$

We are given that $E = 4\sqrt{2}$.

$\dfrac{2\sqrt{2}}{\sin(2\theta)} = 4\sqrt{2} \Rightarrow \sin(2\theta) = \dfrac{1}{2}$

Since $\theta \in \left(0, \dfrac{\pi}{2}\right)$, we have $2\theta \in (0, \pi)$. The solutions for $2\theta$ are $\dfrac{\pi}{6}$ and $\dfrac{5\pi}{6}$.

Thus, $\theta = \dfrac{\pi}{12}, \dfrac{5\pi}{12}$.

The largest element is $\alpha = \dfrac{5\pi}{12}$ and the smallest is $\beta = \dfrac{\pi}{12}$.

The absolute difference is $\alpha - \beta = \dfrac{5\pi}{12} - \dfrac{\pi}{12} = \dfrac{4\pi}{12} = \dfrac{\pi}{3}$.

Answer: $\dfrac{\pi}{3}$