A curve $C$ has parametric equations $$x = 2 \sin 2t, \quad y = \cos 2t, \quad \text{for } 0 \le t < \pi.$$

1. Find the derivatives of the parametric equations with respect to $t$.

$x = 2\sin 2t \implies \frac{dx}{dt} = 4\cos 2t$
$y = \cos 2t \implies \frac{dy}{dt} = -2\sin 2t$

2. Determine the gradient of the normal to the curve at parameter $\theta$.

At point $P$ with parameter $\theta$, the gradient of the tangent is:
$\frac{dy}{dx} = \frac{\frac{dy}{dt}}{\frac{dx}{dt}} = \frac{-2\sin 2\theta}{4\cos 2\theta} = -\frac{\sin 2\theta}{2\cos 2\theta}$

Since the normal is perpendicular to the tangent, its gradient $m_N$ is the negative reciprocal:
$m_N = -\frac{1}{\frac{dy}{dx}} = \frac{2\cos 2\theta}{\sin 2\theta}$

3. Form the equation of the normal and compare it to the given form to find $m$.

The coordinates of $P$ are $(2\sin 2\theta, \cos 2\theta)$.
The equation of the normal is:
$y - \cos 2\theta = \left(\frac{2\cos 2\theta}{\sin 2\theta}\right) (x - 2\sin 2\theta)$

Multiply both sides by $\sin 2\theta$:
$(\sin 2\theta)y - \sin 2\theta\cos 2\theta = (2\cos 2\theta)x - 4\sin 2\theta\cos 2\theta$

Rearrange the terms into the required format:
$(2\cos 2\theta)x - (\sin 2\theta)y = 4\sin 2\theta\cos 2\theta - \sin 2\theta\cos 2\theta$
$(2\cos 2\theta)x - (\sin 2\theta)y = 3\sin 2\theta\cos 2\theta$

Comparing this with $(2\cos 2\theta)x - (\sin 2\theta)y = m\sin 2\theta\cos 2\theta$, we find that:
$m = 3$

1. Determine the value of the integer $m$ from the equation of the normal.

$\frac{\mathrm{d}x}{\mathrm{d}t} = 4\cos 2t$, $\quad \frac{\mathrm{d}y}{\mathrm{d}t} = -2\sin 2t$

Gradient of normal at $P$: $m_N = -\frac{\mathrm{d}x/\mathrm{d}t}{\mathrm{d}y/\mathrm{d}t} = \frac{4\cos 2\theta}{2\sin 2\theta} = \frac{2\cos 2\theta}{\sin 2\theta}$

Equation of normal at $P$:
$y - \cos 2\theta = \frac{2\cos 2\theta}{\sin 2\theta} (x - 2\sin 2\theta)$
$y\sin 2\theta - \sin 2\theta\cos 2\theta = x(2\cos 2\theta) - 4\sin 2\theta\cos 2\theta$
$(2\cos 2\theta)x - (\sin 2\theta)y = 3\sin 2\theta\cos 2\theta$

Comparing this with the given equation, $m = 3$.

2. Find the coordinates of the intercepts $A$ and $B$.

To find $A$, set $y = 0$:
$(2\cos 2\theta)x = 3\sin 2\theta\cos 2\theta \implies x = \frac{3}{2}\sin 2\theta$
Thus, $A = \left(\frac{3}{2}\sin 2\theta, 0\right)$.

To find $B$, set $x = 0$:
$-(\sin 2\theta)y = 3\sin 2\theta\cos 2\theta \implies y = -3\cos 2\theta$
Thus, $B = (0, -3\cos 2\theta)$.

3. Find the parametric coordinates of the midpoint of $AB$.

Let the midpoint of $AB$ be $(x, y)$.
$x = \frac{1}{2}\left(\frac{3}{2}\sin 2\theta + 0\right) = \frac{3}{4}\sin 2\theta$
$y = \frac{1}{2}\left(0 - 3\cos 2\theta\right) = -\frac{3}{2}\cos 2\theta$

4. Eliminate the parameter $\theta$ to find the cartesian equation of the locus.

From the midpoint coordinates:
$\sin 2\theta = \frac{4x}{3} \quad \text{and} \quad \cos 2\theta = -\frac{2y}{3}$

Using the identity $\sin^2 2\theta + \cos^2 2\theta = 1$:
$\left(\frac{4x}{3}\right)^2 + \left(-\frac{2y}{3}\right)^2 = 1$
$\frac{16x^2}{9} + \frac{4y^2}{9} = 1$
$16x^2 + 4y^2 = 9$