(i) To find the equation of the curve, integrate \(\frac{dy}{dx} = 3x^2 - 4x + 1\).

\(y = \int (3x^2 - 4x + 1) \, dx = x^3 - 2x^2 + x + c\).

Use the point (1, 5) to find \(c\):

\(5 = 1^3 - 2(1)^2 + 1 + c\)

\(5 = 1 - 2 + 1 + c\)

\(5 = 0 + c\)

\(c = 5\)

Thus, the equation of the curve is \(y = x^3 - 2x^2 + x + 5\).

(ii) To find where the gradient is positive, solve \(3x^2 - 4x + 1 > 0\).

Factorize or use the quadratic formula to find the roots:

\(3x^2 - 4x + 1 = 0\)

\(x = \frac{-b \pm \sqrt{b^2 - 4ac}}{2a} = \frac{4 \pm \sqrt{(-4)^2 - 4 \cdot 3 \cdot 1}}{2 \cdot 3}\)

\(x = \frac{4 \pm \sqrt{16 - 12}}{6}\)

\(x = \frac{4 \pm 2}{6}\)

\(x = 1\) or \(x = \frac{1}{3}\)

The inequality \(3x^2 - 4x + 1 > 0\) holds for \(x < \frac{1}{3}\) and \(x > 1\).

(i) To find the equation of the curve, integrate the gradient function:

\(\frac{dy}{dx} = \sqrt{1 + 2x} = (1 + 2x)^{1/2}\)

Integrate: \(y = \int (1 + 2x)^{1/2} \, dx\)

Using substitution, let \(u = 1 + 2x\), then \(du = 2 \, dx\), so \(dx = \frac{du}{2}\).

\(y = \int u^{1/2} \cdot \frac{1}{2} \, du = \frac{1}{2} \cdot \frac{u^{3/2}}{3/2} + C = \frac{1}{3}(1 + 2x)^{3/2} + C\)

\(y = \frac{2}{3}(1 + 2x)^{3/2} + C\)

Use the point \((4, 11)\) to find \(C\):

\(11 = \frac{2}{3}(1 + 2 \cdot 4)^{3/2} + C\)

\(11 = \frac{2}{3}(9)^{3/2} + C\)

\(11 = \frac{2}{3} \cdot 27 + C\)

\(11 = 18 + C\)

\(C = 2\)

Thus, the equation of the curve is \(y = \frac{2}{3}(1 + 2x)^{3/2} + 2\).

(ii) To find the y-intercept, set \(x = 0\):

\(y = \frac{2}{3}(1 + 2 \cdot 0)^{3/2} + 2\)

\(y = \frac{2}{3}(1)^{3/2} + 2\)

\(y = \frac{2}{3} + 2\)

\(y = \frac{2}{3} + \frac{6}{3}\)

\(y = \frac{8}{3}\)

The curve intersects the y-axis at \((0, \frac{8}{3})\).

(i) At \(P(1,5)\), the gradient \(m\) of the tangent is \(\frac{4}{3}\). The gradient of the normal is \(-\frac{3}{4}\) (since \(m_1 \cdot m_2 = -1\)). The equation of the normal is \(y - 5 = -\frac{3}{4}(x - 1)\). Setting \(y = 0\) to find where it crosses the x-axis, we solve \(0 - 5 = -\frac{3}{4}(x - 1)\), giving \(x = \frac{23}{3}\). Thus, \(Q \left( \frac{23}{3}, 0 \right)\).

(ii) Integrate \(\frac{dy}{dx} = \frac{12}{(2x+1)^2}\) to find \(y\). Let \(u = 2x+1\), then \(du = 2dx\), so \(dx = \frac{du}{2}\). The integral becomes \(\int \frac{12}{u^2} \cdot \frac{du}{2} = -\frac{6}{u} + C\). Substituting back, \(y = -\frac{6}{2x+1} + C\). Using \(P(1,5)\), \(5 = -\frac{6}{3} + C\), so \(C = 7\). Therefore, \(y = -\frac{6}{2x+1} + 7\).

(iii) Given \(\frac{dx}{dt} = 0.3\), find \(\frac{dy}{dt}\) using \(\frac{dy}{dt} = \frac{dy}{dx} \cdot \frac{dx}{dt}\). At \(x = 1\), \(\frac{dy}{dx} = \frac{4}{3}\). Thus, \(\frac{dy}{dt} = \frac{4}{3} \times 0.3 = 0.4\) units per second.

Given \(\frac{dy}{dx} = \frac{k}{\sqrt{x}}\), integrate to find \(y\):

\(y = \int \frac{k}{\sqrt{x}} \, dx = k \int x^{-1/2} \, dx = k \cdot \frac{x^{1/2}}{1/2} + c = 2k\sqrt{x} + c\).

Substitute the point \(P(1, -1)\):

\(-1 = 2k\sqrt{1} + c \Rightarrow -1 = 2k + c\).

Substitute the point \(Q(4, 4)\):

\(4 = 2k\sqrt{4} + c \Rightarrow 4 = 4k + c\).

Solving the system of equations:

\(-1 = 2k + c\)

\(4 = 4k + c\)

Subtract the first equation from the second:

\(4 - (-1) = 4k - 2k\)

\(5 = 2k \Rightarrow k = \frac{5}{2}\).

Substitute \(k = \frac{5}{2}\) into \(-1 = 2k + c\):

\(-1 = 2 \times \frac{5}{2} + c \Rightarrow -1 = 5 + c \Rightarrow c = -6\).

The equation of the curve is:

\(y = 5\sqrt{x} - 6\).

(i) To find the equation of the curve, integrate \(\frac{dy}{dx} = (5x - 1)^{\frac{1}{2}} - 2\).

Integrate \((5x - 1)^{\frac{1}{2}}\) using substitution:

Let \(u = 5x - 1\), then \(du = 5 \, dx\), so \(dx = \frac{du}{5}\).

\(\int (5x - 1)^{\frac{1}{2}} \, dx = \int u^{\frac{1}{2}} \cdot \frac{du}{5} = \frac{1}{5} \cdot \frac{2}{3} u^{\frac{3}{2}} + C = \frac{2}{15} (5x - 1)^{\frac{3}{2}} + C\).

Integrate \(-2\) to get \(-2x + D\).

Thus, \(y = \frac{2}{15} (5x - 1)^{\frac{3}{2}} - 2x + C\).

Using the point (2, 3):

\(3 = \frac{2}{15} (5 \times 2 - 1)^{\frac{3}{2}} - 2 \times 2 + C\).

\(3 = \frac{2}{15} \times 27 - 4 + C\).

\(3 = \frac{54}{15} - 4 + C\).

\(3 = \frac{18}{5} - 4 + C\).

\(C = 7 - \frac{18}{5}\).

\(C = \frac{17}{5}\).

Thus, \(y = \frac{2(5x-1)^{\frac{3}{2}}}{15} - 2x + \frac{17}{5}\).

(ii) Differentiate \(\frac{dy}{dx} = (5x - 1)^{\frac{1}{2}} - 2\) to find \(\frac{d^2y}{dx^2}\).

\(\frac{d^2y}{dx^2} = \frac{d}{dx}((5x - 1)^{\frac{1}{2}})\).

Using the chain rule: \(\frac{1}{2}(5x-1)^{-\frac{1}{2}} \times 5\).

\(\frac{d^2y}{dx^2} = \frac{5}{2}(5x-1)^{-\frac{1}{2}}\).

(iii) To find the stationary point, set \(\frac{dy}{dx} = 0\).

\((5x - 1)^{\frac{1}{2}} - 2 = 0\).

\((5x - 1)^{\frac{1}{2}} = 2\).

\(5x - 1 = 4\).

\(5x = 5\).

\(x = 1\).

Substitute \(x = 1\) into \(y = \frac{2(5x-1)^{\frac{3}{2}}}{15} - 2x + \frac{17}{5}\):

\(y = \frac{2(5 \times 1 - 1)^{\frac{3}{2}}}{15} - 2 \times 1 + \frac{17}{5}\).

\(y = \frac{2 \times 8}{15} - 2 + \frac{17}{5}\).

\(y = \frac{16}{15} - 2 + \frac{17}{5}\).

\(y = \frac{16}{15} - \frac{30}{15} + \frac{51}{15}\).

\(y = \frac{37}{15}\).

Stationary point is \((1, \frac{37}{15})\).

Check the nature using \(\frac{d^2y}{dx^2}\):

\(\frac{d^2y}{dx^2} = \frac{5}{2}(5 \times 1 - 1)^{-\frac{1}{2}} = \frac{5}{2} \times \frac{1}{2} = \frac{5}{4}\).

Since \(\frac{d^2y}{dx^2} > 0\), it is a minimum point.