The curve $C$ has equation $y = \frac{x - 2}{kx^2 + x - 2}$, where $k > 1$.

1. Find the coordinates of point $A$ where the curve cuts the $y$-axis.

Substitute $x = 0$ into the equation of $C$:
$y = \frac{0 - 2}{k(0)^2 + 0 - 2} = \frac{-2}{-2} = 1$
Point $A$ is $(0, 1)$.

2. Find the gradient of the tangent at $A$ by finding $\frac{\mathrm{d}y}{\mathrm{d}x}$ and evaluating it at $x = 0$.

Using the quotient rule:
$\frac{\mathrm{d}y}{\mathrm{d}x} = \frac{(kx^2 + x - 2)\frac{\mathrm{d}}{\mathrm{d}x}(x - 2) - (x - 2)\frac{\mathrm{d}}{\mathrm{d}x}(kx^2 + x - 2)}{(kx^2 + x - 2)^2}$
$\frac{\mathrm{d}y}{\mathrm{d}x} = \frac{(kx^2 + x - 2)(1) - (x - 2)(2kx + 1)}{(kx^2 + x - 2)^2}$

Substitute $x = 0$:
$\left. \frac{\mathrm{d}y}{\mathrm{d}x} \right|_{x=0} = \frac{(-2)(1) - (-2)(1)}{(-2)^2} = \frac{-2 + 2}{4} = 0$
The gradient of the tangent at $A$ is $0$.

3. State the equation of the tangent at point $A$.

Since the gradient is $0$ and the line passes through $(0, 1)$, the tangent is a horizontal line.
$y - 1 = 0(x - 0)$
$y = 1$

1. Determine the axial intercepts of the curve.

To find the $x$-intercept, set $y = 0$:
$\frac{x - 2}{kx^2 + x - 2} = 0 \implies x - 2 = 0 \implies x = 2$
Thus, the $x$-intercept is $(2, 0)$.

To find the $y$-intercept, set $x = 0$:
$y = \frac{0 - 2}{k(0)^2 + 0 - 2} = \frac{-2}{-2} = 1$
Thus, the $y$-intercept is $(0, 1)$.

2. Find the equations of the horizontal and vertical asymptotes.

For the horizontal asymptote, we consider the behavior as $x \to \pm\infty$:
$\lim_{x \to \pm\infty} \frac{x - 2}{kx^2 + x - 2} = \lim_{x \to \pm\infty} \frac{\frac{1}{x} - \frac{2}{x^2}}{k + \frac{1}{x} - \frac{2}{x^2}} = 0$
So, the horizontal asymptote is $y = 0$.

For vertical asymptotes, the denominator approaches zero:
$kx^2 + x - 2 = 0$
Using the quadratic formula to solve for $x$:
$x = \frac{-1 \pm \sqrt{1^2 - 4(k)(-2)}}{2k} = \frac{-1 \pm \sqrt{1 + 8k}}{2k}$
These are the equations of the vertical asymptotes.

3. Differentiate the equation to find the $x$-coordinates of the turning points.

Using the quotient rule $\frac{v u' - u v'}{v^2}$:
$\frac{dy}{dx} = \frac{(kx^2 + x - 2)(1) - (x - 2)(2kx + 1)}{(kx^2 + x - 2)^2}$
$\frac{dy}{dx} = \frac{kx^2 + x - 2 - (2kx^2 + x - 4kx - 2)}{(kx^2 + x - 2)^2}$
$\frac{dy}{dx} = \frac{-kx^2 + 4kx}{(kx^2 + x - 2)^2}$
For turning points, set $\frac{dy}{dx} = 0$:
$-kx^2 + 4kx = 0 \implies -kx(x - 4) = 0$
Since $k > 1$ (meaning $k \neq 0$), the possible solutions are:
$x = 0 \quad \text{or} \quad x = 4$

4. Calculate the corresponding $y$-coordinates to finalize the turning points.

When $x = 0$:
$y = 1$
When $x = 4$:
$y = \frac{4 - 2}{k(4^2) + 4 - 2} = \frac{2}{16k + 2} = \frac{1}{8k + 1}$
So, the coordinates of the turning points are $(0, 1)$ and $\left(4, \frac{1}{8k + 1}\right)$.

(The student would then synthesize these features to sketch the graph, with $(0, 1)$ as a local minimum and $\left(4, \frac{1}{8k+1}\right)$ as a local maximum, properly bound by the respective asymptotes).

1. Find the coordinates of point $B$ where the curve intersects the $x$-axis.

At the $x$-axis, $y = 0$.
$\frac{x - 2}{kx^2 + x - 2} = 0$
$x - 2 = 0 \implies x = 2$
The coordinates of $B$ are $(2, 0)$.

2. Differentiate $y$ with respect to $x$ to find the gradient function.

Using the quotient rule:
$\frac{\mathrm{d}y}{\mathrm{d}x} = \frac{(kx^2 + x - 2)\frac{\mathrm{d}}{\mathrm{d}x}(x - 2) - (x - 2)\frac{\mathrm{d}}{\mathrm{d}x}(kx^2 + x - 2)}{(kx^2 + x - 2)^2}$
$\frac{\mathrm{d}y}{\mathrm{d}x} = \frac{(kx^2 + x - 2)(1) - (x - 2)(2kx + 1)}{(kx^2 + x - 2)^2}$

3. Evaluate the gradient of the normal at $B$ and find its equation.

Substitute $x = 2$ into the derivative to find the gradient of the tangent, $m_t$:
$m_t = \frac{(k(2)^2 + 2 - 2)(1) - (0)(2k(2) + 1)}{(k(2)^2 + 2 - 2)^2}$
$m_t = \frac{4k}{(4k)^2} = \frac{1}{4k}$
The gradient of the normal, $m_n$, is $-\frac{1}{m_t}$:
$m_n = -4k$
Using the point-gradient form for $B(2, 0)$:
$y - 0 = -4k(x - 2)$
$y = -4kx + 8k$

1. State or determine the equations of the tangent at $A$ and the normal at $B$ from earlier parts.

From earlier parts, $C$ cuts the $y$-axis at $A(0,1)$ and the $x$-axis at $B(2,0)$.
$\frac{\mathrm{d}y}{\mathrm{d}x} = \frac{(kx^2+x-2)(1) - (x-2)(2kx+1)}{(kx^2+x-2)^2}$
At $A(0,1)$, $\frac{\mathrm{d}y}{\mathrm{d}x} = 0$, so the equation of the tangent is $y = 1$.
At $B(2,0)$, $\frac{\mathrm{d}y}{\mathrm{d}x} = \frac{1(4k) - 0}{(4k)^2} = \frac{1}{4k}$. The gradient of the normal is $-4k$.
Equation of the normal at $B$: $y - 0 = -4k(x - 2) \implies y = -4kx + 8k$.

2. Find the point of intersection between the tangent at $A$ and the normal at $B$.

Substitute $y = 1$ into the normal equation:
$1 = -4kx + 8k$
$4kx = 8k - 1 \implies x = \frac{8k-1}{4k} = 2 - \frac{1}{4k}$
The intersection point is $\left(2 - \frac{1}{4k}, 1\right)$.

3. Calculate the area of the region bounded by the tangent, the normal, and both axes.

The region is bounded by $y=1$, $y=-4kx+8k$, $y=0$ ($x$-axis), and $x=0$ ($y$-axis).
This forms a right-angled trapezium with vertices $(0,0)$, $(2,0)$, $\left(2 - \frac{1}{4k}, 1\right)$, and $(0,1)$.
The parallel horizontal sides lie on $y=0$ and $y=1$ with lengths $2$ and $2 - \frac{1}{4k}$ respectively. The height along the $y$-axis is $1$.
$\text{Area} = \frac{1}{2} \times \text{height} \times (\text{sum of parallel sides})$
$\text{Area} = \frac{1}{2} (1) \left[ 2 + \left(2 - \frac{1}{4k}\right) \right] = \frac{1}{2} \left(4 - \frac{1}{4k}\right) = 2 - \frac{1}{8k}$

4. Use the given condition $k > 1$ to show that the area is strictly greater than $\frac{15}{8}$.

We are given that $k > 1$.
$8k > 8 \implies \frac{1}{8k} < \frac{1}{8}$
Multiplying by $-1$ reverses the inequality:
$-\frac{1}{8k} > -\frac{1}{8}$
Therefore, adding $2$ to both sides:
$\text{Area} = 2 - \frac{1}{8k} > 2 - \frac{1}{8} = \frac{15}{8}$
Thus, the area is more than $\frac{15}{8}$ square units.