To find the stationary point, we need to differentiate the function \(y = \cos^3 x \sqrt{\sin x}\) and set the derivative equal to zero.

Using the product rule and chain rule, the derivative is:

\(\frac{dy}{dx} = -3\cos^2 x \sin x \sqrt{\sin x} + \frac{\cos^3 x \cos x}{2\sqrt{\sin x}}\).

Simplifying, we get:

\(-3\cos^2 x \sin^2 x + \frac{1}{2}\cos^4 x = 0\).

Rearranging gives:

\(7\cos^3 x = 6\).

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

\(x = 0.388\).

(a) To find the stationary point, we first find the derivative of \(y = xe^{1-2x}\) using the product rule. Let \(u = x\) and \(v = e^{1-2x}\). Then \(\frac{du}{dx} = 1\) and \(\frac{dv}{dx} = -2e^{1-2x}\).

Using the product rule, \(\frac{dy}{dx} = u \frac{dv}{dx} + v \frac{du}{dx} = x(-2e^{1-2x}) + e^{1-2x}\).

Simplifying, \(\frac{dy}{dx} = e^{1-2x} - 2xe^{1-2x} = e^{1-2x}(1 - 2x)\).

Set \(\frac{dy}{dx} = 0\) to find the stationary point: \(e^{1-2x}(1 - 2x) = 0\).

Since \(e^{1-2x} \neq 0\), we have \(1 - 2x = 0\), giving \(x = \frac{1}{2}\).

Substitute \(x = \frac{1}{2}\) back into the original equation to find \(y\):

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

Thus, the coordinates of the stationary point are \(\left( \frac{1}{2}, \frac{1}{2} \right)\).

(b) To determine the nature of the stationary point, we find the second derivative \(\frac{d^2y}{dx^2}\).

Differentiate \(\frac{dy}{dx} = e^{1-2x}(1 - 2x)\) again using the product rule:

\(\frac{d^2y}{dx^2} = \frac{d}{dx}(e^{1-2x}(1 - 2x)) = -2e^{1-2x}(1 - 2x) - 2e^{1-2x}\).

At \(x = \frac{1}{2}\), \(\frac{d^2y}{dx^2} = -2e^{1-2(\frac{1}{2})}(1 - 2(\frac{1}{2})) - 2e^{1-2(\frac{1}{2})} = -2e^{0}(0) - 2e^{0} = -2\).

Since \(\frac{d^2y}{dx^2} < 0\), the stationary point is a maximum.

To find the maximum point \(M\) of the curve \(y = \frac{\ln x}{x^4}\), we first find the derivative \(y'\) using the quotient rule.

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

Here, \(u = \ln x\) and \(v = x^4\).

Thus, \(u' = \frac{1}{x}\) and \(v' = 4x^3\).

Applying the quotient rule:

\(y' = \frac{\left(\frac{1}{x}\right)x^4 - (\ln x)(4x^3)}{(x^4)^2} = \frac{x^3 - 4x^3 \ln x}{x^8} = \frac{x^3(1 - 4 \ln x)}{x^8} = \frac{1 - 4 \ln x}{x^5}\).

To find the maximum, set \(y' = 0\):

\(\frac{1 - 4 \ln x}{x^5} = 0\).

This implies \(1 - 4 \ln x = 0\).

Solving for \(x\), we get \(\ln x = \frac{1}{4}\).

Thus, \(x = e^{1/4} = \sqrt[4]{e}\).

Substitute \(x = \sqrt[4]{e}\) back into the original equation to find \(y\):

\(y = \frac{\ln(\sqrt[4]{e})}{(\sqrt[4]{e})^4} = \frac{\frac{1}{4}}{e} = \frac{1}{4e}\).

Therefore, the coordinates of \(M\) are \(\left( \sqrt[4]{e}, \frac{1}{4e} \right)\).

To find the stationary points, we need to find where the derivative \(\frac{dy}{dx} = 0\).

Using the product rule, the derivative is:

\(\frac{dy}{dx} = -5e^{-5x} \tan^2 x + 2e^{-5x} \tan x \sec^2 x\).

Setting \(\frac{dy}{dx} = 0\), we have:

\(-5e^{-5x} \tan^2 x + 2e^{-5x} \tan x \sec^2 x = 0\).

Factor out \(e^{-5x} \tan x\):

\(e^{-5x} \tan x (-5 \tan x + 2 \sec^2 x) = 0\).

Since \(e^{-5x} \neq 0\) and \(\tan x = 0\) gives \(x = 0\), we solve:

\(-5 \tan x + 2 \sec^2 x = 0\).

Using \(\sec^2 x = 1 + \tan^2 x\), we get:

\(-5 \tan x + 2(1 + \tan^2 x) = 0\).

Simplifying, we have:

\(2 \tan^2 x - 5 \tan x + 2 = 0\).

Solving this quadratic equation in \(\tan x\), we find:

\(\tan x = 0.546\) or \(\tan x = 1.107\).

Thus, \(x = \arctan(0.546) \approx 0.464\) and \(x = \arctan(1.107) \approx 1.107\).

Therefore, the \(x\)-coordinates of the stationary points are \(x = 0, 0.464, 1.107\).

To find the stationary point, we need to find where the derivative \(y'\) is zero.

Given \(y = x^{-\frac{2}{3}} \ln x\), we use the product rule to differentiate:

Let \(u = x^{-\frac{2}{3}}\) and \(v = \ln x\).

Then \(u' = -\frac{2}{3}x^{-\frac{5}{3}}\) and \(v' = \frac{1}{x}\).

Using the product rule, \(y' = u'v + uv'\).

So, \(y' = -\frac{2}{3}x^{-\frac{5}{3}} \ln x + x^{-\frac{2}{3}} \cdot \frac{1}{x}\).

Simplifying, \(y' = -\frac{2}{3}x^{-\frac{5}{3}} \ln x + x^{-\frac{5}{3}}\).

Factor out \(x^{-\frac{5}{3}}\):

\(y' = x^{-\frac{5}{3}} \left( -\frac{2}{3} \ln x + 1 \right)\).

Set \(y' = 0\):

\(x^{-\frac{5}{3}} \left( -\frac{2}{3} \ln x + 1 \right) = 0\).

Since \(x^{-\frac{5}{3}} \neq 0\) for \(x > 0\), solve:

\(-\frac{2}{3} \ln x + 1 = 0\).

\(\ln x = \frac{3}{2}\).

\(x = e^{\frac{3}{2}}\).

Substitute \(x = e^{\frac{3}{2}}\) back into the original equation to find \(y\):

\(y = \left( e^{\frac{3}{2}} \right)^{-\frac{2}{3}} \ln \left( e^{\frac{3}{2}} \right)\).

\(y = e^{-1} \cdot \frac{3}{2}\).

\(y = \frac{3}{2e}\).

Thus, the exact coordinates of the stationary point are \(\left( e^{\frac{3}{2}}, \frac{3}{2e} \right)\).