Given the equation \(\sin^{-1}(4x^4 + x^2) = \frac{1}{6}\pi\), we first take the sine of both sides to get:

\(4x^4 + x^2 = \sin\left(\frac{1}{6}\pi\right)\).

Since \(\sin\left(\frac{1}{6}\pi\right) = \frac{1}{2}\), we have:

\(4x^4 + x^2 = \frac{1}{2}\).

Let \(y = x^2\), then the equation becomes:

\(4y^2 + y - \frac{1}{2} = 0\).

This is a quadratic equation in \(y\). Solving for \(y\) using the quadratic formula:

\(y = \frac{-b \pm \sqrt{b^2 - 4ac}}{2a}\), where \(a = 4\), \(b = 1\), \(c = -\frac{1}{2}\).

\(y = \frac{-1 \pm \sqrt{1 + 8}}{8}\).

\(y = \frac{-1 \pm 3}{8}\).

This gives \(y = \frac{1}{4}\) or \(y = -\frac{1}{2}\).

Since \(y = x^2\) and \(y\) must be non-negative, we have \(x^2 = \frac{1}{4}\).

Thus, \(x = \pm \frac{1}{2}\).

Given \(\alpha = \cos^{-1}\left(\frac{8}{17}\right)\), we have \(\cos \alpha = \frac{8}{17}\).

Using the identity \(\sin^2 \alpha + \cos^2 \alpha = 1\), we find \(\sin \alpha = \sqrt{1 - \left(\frac{8}{17}\right)^2}\).

Calculate \(\sin \alpha = \sqrt{1 - \frac{64}{289}} = \sqrt{\frac{225}{289}} = \frac{15}{17}\).

Now, \(\tan \alpha = \frac{\sin \alpha}{\cos \alpha} = \frac{15/17}{8/17} = \frac{15}{8}\).

We need to find \(\frac{1}{\sin \alpha} + \frac{1}{\tan \alpha}\).

\(\frac{1}{\sin \alpha} = \frac{17}{15}\) and \(\frac{1}{\tan \alpha} = \frac{8}{15}\).

Adding these, \(\frac{1}{\sin \alpha} + \frac{1}{\tan \alpha} = \frac{17}{15} + \frac{8}{15} = \frac{25}{15} = \frac{5}{3}\).

(i) The maximum height occurs when \(\cos kt = -1\). Substituting this into the formula gives:

\(h = 60(1 - (-1)) = 60 \times 2 = 120 \text{ m}\)

(ii) Given that one complete revolution takes 30 minutes, \(kt = 2\pi\) when \(t = 30\). Therefore,

\(30k = 2\pi\)

\(k = \frac{2\pi}{30} = \frac{\pi}{15}\)

(iii) To find the time when the height is above 90 m, set \(h = 90\):

\(90 = 60(1 - \cos kt)\)

\(1 - \cos kt = \frac{90}{60} = 1.5\)

\(\cos kt = -0.5\)

The solutions for \(\cos kt = -0.5\) are \(kt = \frac{2\pi}{3}\) or \(kt = \frac{4\pi}{3}\).

Solving for \(t\):

\(t = \frac{2\pi}{3k} = \frac{2\pi}{3 \times \frac{\pi}{15}} = 10 \text{ minutes}\)

\(t = \frac{4\pi}{3k} = \frac{4\pi}{3 \times \frac{\pi}{15}} = 20 \text{ minutes}\)

Thus, the passenger is above 90 m for 10 minutes.

Since \(\theta\) is an obtuse angle, \(\theta\) is in the second quadrant where sine is positive and cosine is negative.

(i) Using the identity \(\sin^2 \theta + \cos^2 \theta = 1\), we have:

\(\cos^2 \theta = 1 - \sin^2 \theta = 1 - k^2\)

\(\cos \theta = -\sqrt{1-k^2}\) because cosine is negative in the second quadrant.

(ii) \(\tan \theta = \frac{\sin \theta}{\cos \theta} = \frac{k}{-\sqrt{1-k^2}} = -\frac{k}{\sqrt{1-k^2}}\)

(iii) \(\sin(\theta + \pi) = -\sin \theta = -k\) because adding \(\pi\) to an angle reflects it across the origin, changing the sign of sine.

First, calculate \(\arctan(3)\). This gives \(\arctan(3) = 1.249\) radians.

Next, we need to find \(\sin(1.249)\). Using a calculator, \(\sin(1.249) \approx 0.949\).

Set \(x - 1 = 0.949\), so \(x = 0.949 + 1 = 1.949\).

Therefore, the value of \(x\) is approximately 1.95.