(a) To find the stationary points, we first find the derivative \(\frac{dy}{dx}\).

\(\frac{dy}{dx} = \left( \frac{5}{3} (4x - 3)^{\frac{2}{3}} \right) \times 4 - \frac{20}{3}\).

Setting \(\frac{dy}{dx} = 0\) gives:

\(\frac{20}{3} (4x - 3)^{\frac{2}{3}} = \frac{20}{3}\).

This leads to \((4x - 3)^{\frac{2}{3}} = 1\), so \(4x - 3 = \pm 1\).

Solving gives \(x = \frac{1}{2}\) and \(x = 1\).

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

\(\frac{d^2y}{dx^2} = \frac{40}{9} (4x - 3)^{-\frac{1}{3}} \times 4\).

For \(x = \frac{1}{2}\), \(\frac{d^2y}{dx^2} < 0\), so it is a maximum.

For \(x = 1\), \(\frac{d^2y}{dx^2} > 0\), so it is a minimum.

(b) The function \(f\) is increasing where \(\frac{dy}{dx} > 0\).

This occurs when \(x < \frac{1}{2}\) or \(x > 1\).

First, find the derivative \(\frac{dy}{dx}\) of the curve \(y = 2 + \sqrt{25 - x^2}\).

Using the chain rule, \(\frac{dy}{dx} = \frac{1}{2}(25 - x^2)^{-1/2} \times (-2x)\).

Simplifying, \(\frac{dy}{dx} = \frac{-x}{\sqrt{25 - x^2}}\).

Set \(\frac{dy}{dx} = \frac{4}{3}\) and solve for \(x\):

\(\frac{-x}{\sqrt{25 - x^2}} = \frac{4}{3}\).

Square both sides: \(\frac{x^2}{25 - x^2} = \frac{16}{9}\).

Cross-multiply: \(16(25 - x^2) = 9x^2\).

Expand and simplify: \(400 - 16x^2 = 9x^2\).

Combine terms: \(25x^2 = 400\).

Solve for \(x\): \(x^2 = 16\), so \(x = \pm 4\).

Substitute \(x = -4\) into the original equation to find \(y\):

\(y = 2 + \sqrt{25 - (-4)^2} = 2 + \sqrt{9} = 2 + 3 = 5\).

Thus, the coordinates are \((-4, 5)\).

(a) Differentiate \(y = 54x - (2x - 7)^3\) with respect to \(x\):

\(\frac{dy}{dx} = 54 - 3(2x - 7)^2 \cdot 2\)

\(\frac{dy}{dx} = 54 - 6(2x - 7)^2\)

Differentiate again to find \(\frac{d^2y}{dx^2}\):

\(\frac{d^2y}{dx^2} = -6 \cdot 2(2x - 7) \cdot 2\)

\(\frac{d^2y}{dx^2} = -24(2x - 7)\)

(b) Set \(\frac{dy}{dx} = 0\):

\(54 - 6(2x - 7)^2 = 0\)

\(6(2x - 7)^2 = 54\)

\((2x - 7)^2 = 9\)

\(2x - 7 = \pm 3\)

\(2x = 10\) or \(2x = 4\)

\(x = 5\) or \(x = 2\)

For \(x = 5\), \(y = 54(5) - (2(5) - 7)^3 = 270 - 27 = 243\)

For \(x = 2\), \(y = 54(2) - (2(2) - 7)^3 = 108 - (-3)^3 = 135\)

(c) Evaluate \(\frac{d^2y}{dx^2}\) at the stationary points:

For \(x = 5\), \(\frac{d^2y}{dx^2} = -24(3) = -72\), indicating a maximum.

For \(x = 2\), \(\frac{d^2y}{dx^2} = -24(-3) = 72\), indicating a minimum.

(a) Differentiate \(y = (3 - 2x)^3 + 24x\) with respect to \(x\):

\(\frac{dy}{dx} = 3(3 - 2x)^2(-2) + 24 = -6(3 - 2x)^2 + 24\).

Differentiate again to find \(\frac{d^2y}{dx^2}\):

\(\frac{d^2y}{dx^2} = -12(3 - 2x)(-2) = 24(3 - 2x)\).

(b) Set \(\frac{dy}{dx} = 0\):

\(-6(3 - 2x)^2 + 24 = 0\)

\(6(3 - 2x)^2 = 24\)

\((3 - 2x)^2 = 4\)

\(3 - 2x = \pm 2\)

\(x = \frac{1}{2}\) or \(x = \frac{5}{2}\)

For \(x = \frac{1}{2}\), \(y = (3 - 2 \times \frac{1}{2})^3 + 24 \times \frac{1}{2} = 20\)

For \(x = \frac{5}{2}\), \(y = (3 - 2 \times \frac{5}{2})^3 + 24 \times \frac{5}{2} = 52\)

(c) Evaluate \(\frac{d^2y}{dx^2}\) at the stationary points:

For \(x = \frac{1}{2}\), \(\frac{d^2y}{dx^2} = 24(3 - 2 \times \frac{1}{2}) = 48\), indicating a minimum.

For \(x = \frac{5}{2}\), \(\frac{d^2y}{dx^2} = 24(3 - 2 \times \frac{5}{2}) = -48\), indicating a maximum.

To find the stationary points, we first find the derivative of the curve:

\(\frac{dy}{dx} = 3x^2 + 2x - 8\).

Set \(\frac{dy}{dx} = 0\) to find the stationary points:

\(3x^2 + 2x - 8 = 0\).

Solving this quadratic equation using the quadratic formula \(x = \frac{-b \pm \sqrt{b^2 - 4ac}}{2a}\), where \(a = 3\), \(b = 2\), and \(c = -8\), we find:

\(x = \frac{-2 \pm \sqrt{2^2 - 4 \times 3 \times (-8)}}{2 \times 3}\).

\(x = \frac{-2 \pm \sqrt{4 + 96}}{6}\).

\(x = \frac{-2 \pm \sqrt{100}}{6}\).

\(x = \frac{-2 \pm 10}{6}\).

This gives the solutions \(x = \frac{8}{6} = \frac{4}{3}\) and \(x = \frac{-12}{6} = -2\).

For the curve to have no stationary points in the interval \(a < x < b\), the interval must not include these solutions. Therefore, the least possible value of \(a\) is \(-2\) and the greatest possible value of \(b\) is \(\frac{4}{3}\).