To find the stationary points, we need to find where the derivative \(\frac{dy}{dx}\) is zero.

Using the product rule, the derivative of \(y = \cos x \sin 2x\) is:

\(\frac{dy}{dx} = -\sin x \sin 2x + 2\cos x \cos 2x\).

Using the double angle formula \(\cos 2x = 1 - 2\sin^2 x\), express the derivative in terms of \(\sin x\) and \(\cos x\):

\(\frac{dy}{dx} = -\sin x \sin 2x + 2\cos x (1 - 2\sin^2 x)\).

Equate the derivative to zero:

\(-\sin x \sin 2x + 2\cos x \cos 2x = 0\).

Simplify to obtain an equation in one trigonometric function:

\(3\sin 2x = 1\) or \(3\cos 2x = 2\) or \(2\tan 2x = 1\).

Solving \(2\tan 2x = 1\) gives:

\(\tan 2x = \frac{1}{2}\).

Solving for \(x\) in the interval \(0 < x < \frac{1}{2} \pi\), we find:

\(x = 0.615\).

(a) To find \(\frac{dy}{dx}\), use the product rule. Let \(u = x\) and \(v = \arctan\left(\frac{1}{2}x\right)\).

Then \(\frac{du}{dx} = 1\) and \(\frac{dv}{dx} = \frac{1}{1 + \left(\frac{1}{2}x\right)^2} \cdot \frac{1}{2} = \frac{1}{4 + x^2}\).

Using the product rule: \(\frac{dy}{dx} = u \frac{dv}{dx} + v \frac{du}{dx} = x \cdot \frac{1}{4 + x^2} + \arctan\left(\frac{1}{2}x\right)\).

Thus, \(\frac{dy}{dx} = \arctan\left(\frac{1}{2}x\right) + \frac{2x}{x^2 + 4}\).

(b) At \(x = 2\), the y-coordinate is \(y = 2 \arctan(1) = 2 \cdot \frac{\pi}{4} = \frac{\pi}{2}\).

The gradient at \(x = 2\) is \(\frac{dy}{dx} = \arctan(1) + \frac{4}{8} = \frac{\pi}{4} + \frac{1}{2}\).

The equation of the tangent is \(y - \frac{\pi}{2} = \left(\frac{\pi}{4} + \frac{1}{2}\right)(x - 2)\).

Setting \(x = 0\) to find \(p\):

\(p - \frac{\pi}{2} = \left(\frac{\pi}{4} + \frac{1}{2}\right)(-2)\)

\(p = \frac{\pi}{2} - \left(\frac{\pi}{2} + 1\right) = -1\).

To find the equation of the tangent, we first need to find the derivative of the curve \(y = \frac{2 - \tan x}{1 + \tan x}\) using the quotient rule.

The quotient rule states that if \(y = \frac{u}{v}\), then \(y' = \frac{v u' - u v'}{v^2}\).

Let \(u = 2 - \tan x\) and \(v = 1 + \tan x\).

Then \(u' = -\sec^2 x\) and \(v' = \sec^2 x\).

Applying the quotient rule:

\(y' = \frac{(1 + \tan x)(-\sec^2 x) - (2 - \tan x)(\sec^2 x)}{(1 + \tan x)^2}\)

\(= \frac{-(1 + \tan x)\sec^2 x - (2 - \tan x)\sec^2 x}{(1 + \tan x)^2}\)

\(= \frac{-\sec^2 x - \tan x \sec^2 x - 2\sec^2 x + \tan x \sec^2 x}{(1 + \tan x)^2}\)

\(= \frac{-3\sec^2 x}{(1 + \tan x)^2}\)

Now, substitute \(x = \frac{1}{4} \pi\) to find the gradient of the tangent:

\(\tan \left( \frac{1}{4} \pi \right) = 1\) and \(\sec^2 \left( \frac{1}{4} \pi \right) = 2\).

\(y' = \frac{-3 \times 2}{(1 + 1)^2} = \frac{-6}{4} = -\frac{3}{2}\).

The gradient of the tangent is \(-\frac{3}{2}\).

To find the y-intercept \(c\), we need the point on the curve at \(x = \frac{1}{4} \pi\):

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

Using the point-slope form \(y - y_1 = m(x - x_1)\), where \(m = -\frac{3}{2}\), \(x_1 = \frac{1}{4} \pi\), and \(y_1 = \frac{1}{2}\):

\(y - \frac{1}{2} = -\frac{3}{2}(x - \frac{1}{4} \pi)\)

\(y = -\frac{3}{2}x + \frac{3}{8} \pi + \frac{1}{2}\)

Approximating \(\frac{3}{8} \pi\) to 3 significant figures gives \(1.18\).

Thus, \(y = -\frac{3}{2}x + 1.68\).

To find the equation of the tangent, we first need to find the derivative of \(y = x \sin 2x\) using the product rule.

The product rule states that if \(y = u \cdot v\), then \(\frac{dy}{dx} = u'v + uv'\).

Let \(u = x\) and \(v = \sin 2x\).

Then \(u' = 1\) and \(v' = 2 \cos 2x\) (using the chain rule).

So, \(\frac{dy}{dx} = 1 \cdot \sin 2x + x \cdot 2 \cos 2x = \sin 2x + 2x \cos 2x\).

Now, evaluate the derivative at \(x = \frac{1}{4} \pi\):

\(\frac{dy}{dx} \bigg|_{x = \frac{1}{4} \pi} = \sin \left( \frac{1}{2} \pi \right) + 2 \cdot \frac{1}{4} \pi \cdot \cos \left( \frac{1}{2} \pi \right)\).

Since \(\sin \left( \frac{1}{2} \pi \right) = 1\) and \(\cos \left( \frac{1}{2} \pi \right) = 0\), we have:

\(\frac{dy}{dx} \bigg|_{x = \frac{1}{4} \pi} = 1 + 0 = 1\).

The slope of the tangent is 1.

Now, find the point on the curve at \(x = \frac{1}{4} \pi\):

\(y = \frac{1}{4} \pi \sin \left( \frac{1}{2} \pi \right) = \frac{1}{4} \pi \cdot 1 = \frac{1}{4} \pi\).

The point is \(\left( \frac{1}{4} \pi, \frac{1}{4} \pi \right)\).

The equation of the tangent line is:

\(y - \frac{1}{4} \pi = 1 \left( x - \frac{1}{4} \pi \right)\).

Simplifying gives:

\(y = x\).

To differentiate \(\sec x = \frac{1}{\cos x}\), use the quotient rule: \(\frac{d}{dx} \left( \frac{u}{v} \right) = \frac{v \frac{du}{dx} - u \frac{dv}{dx}}{v^2}\).

Let \(u = 1\) and \(v = \cos x\). Then \(\frac{du}{dx} = 0\) and \(\frac{dv}{dx} = -\sin x\).

Applying the quotient rule:

\(\frac{d}{dx} \left( \frac{1}{\cos x} \right) = \frac{\cos x \cdot 0 - 1 \cdot (-\sin x)}{\cos^2 x} = \frac{\sin x}{\cos^2 x} = \frac{\sin x}{\cos x} \cdot \frac{1}{\cos x} = \tan x \sec x.\)

Thus, the derivative of \(\sec x\) is \(\sec x \tan x\).

Now, differentiate \(y = \ln(\sec x + \tan x)\) using the chain rule:

Let \(u = \sec x + \tan x\), then \(y = \ln u\).

\(\frac{dy}{du} = \frac{1}{u}\) and \(\frac{du}{dx} = \sec x \tan x + \sec^2 x\).

Thus, \(\frac{dy}{dx} = \frac{dy}{du} \cdot \frac{du}{dx} = \frac{1}{\sec x + \tan x} \cdot (\sec x \tan x + \sec^2 x)\).

Simplifying:

\(\frac{dy}{dx} = \frac{\sec x (\tan x + \sec x)}{\sec x + \tan x} = \sec x\).