The curve $C$ has parametric equations $x = \sin \theta + 1 \quad \text{and} \quad y = \sqrt{3} \cos \theta - 1, \text{ where } -\frac{\pi}{2} \le \theta \le \frac{\pi}{2}.$

1. Find the derivatives of $x$ and $y$ with respect to the parameter $\theta$.

$x = \sin \theta + 1 \implies \frac{\mathrm{d}x}{\mathrm{d}\theta} = \cos \theta$

$y = \sqrt{3} \cos \theta - 1 \implies \frac{\mathrm{d}y}{\mathrm{d}\theta} = -\sqrt{3} \sin \theta$

2. Apply the chain rule for parametric differentiation to deduce $\frac{\mathrm{d}y}{\mathrm{d}x}$.

$\frac{\mathrm{d}y}{\mathrm{d}x} = \frac{\frac{\mathrm{d}y}{\mathrm{d}\theta}}{\frac{\mathrm{d}x}{\mathrm{d}\theta}}$

$\frac{\mathrm{d}y}{\mathrm{d}x} = \frac{-\sqrt{3} \sin \theta}{\cos \theta}$

$\frac{\mathrm{d}y}{\mathrm{d}x} = -\sqrt{3} \tan \theta$

1. Find the value of parameter $t$ at point $T$ using the given angle of the tangent.

The gradient of the tangent is given by $m_t = \tan\left(\frac{3\pi}{4}\right) = -1$.
From (i), we know $\frac{\mathrm{d}y}{\mathrm{d}x} = -\sqrt{3} \tan \theta$. Equating this to the tangent gradient at $T$:
$-\sqrt{3} \tan t = -1$
$\tan t = \frac{1}{\sqrt{3}}$
Given the domain $-\frac{\pi}{2} \le t \le \frac{\pi}{2}$, we have $t = \frac{\pi}{6}$.

2. Calculate the exact coordinates of $T$.

Substitute $t = \frac{\pi}{6}$ into the parametric equations:
$x = \sin\left(\frac{\pi}{6}\right) + 1 = \frac{1}{2} + 1 = \frac{3}{2}$
$y = \sqrt{3} \cos\left(\frac{\pi}{6}\right) - 1 = \sqrt{3}\left(\frac{\sqrt{3}}{2}\right) - 1 = \frac{3}{2} - 1 = \frac{1}{2}$
Hence, $T$ is the point $\left(\frac{3}{2}, \frac{1}{2}\right)$.

3. Determine the gradient of the normal and formulate its equation.

The gradient of the normal is $m_n = -\frac{1}{m_t} = -\frac{1}{-1} = 1$.
The equation of the normal at $T$ is:
$y - \frac{1}{2} = 1\left(x - \frac{3}{2}\right)$
$y = x - \frac{3}{2} + \frac{1}{2}$
$y = x - 1$

1. Set $\frac{\mathrm{d}y}{\mathrm{d}x} = 0$ to find the value of $\theta$ where the tangent is parallel to the $x$-axis.

For the tangent to be parallel to the $x$-axis, its gradient must be zero:
$\frac{\mathrm{d}y}{\mathrm{d}x} = 0$
From part (i), we have:
$-\sqrt{3} \tan \theta = 0$
$\tan \theta = 0$
Given the restricted domain $-\frac{\pi}{2} \le \theta \le \frac{\pi}{2}$, the only solution is:
$\theta = 0$

2. Substitute $\theta = 0$ into the parametric equation for $y$ to find the $y$-coordinate.

Substitute $\theta = 0$ into $y = \sqrt{3} \cos \theta - 1$:
$y = \sqrt{3} \cos(0) - 1$
Since $\cos(0) = 1$, we get:
$y = \sqrt{3}(1) - 1 = \sqrt{3} - 1$

1. Concisely determine the equations of the normal at $T$ and the tangent from (iii).

From (ii), the tangent makes an angle of $\frac{3\pi}{4}$ with the positive $x$-axis, so its gradient is $\tan\left(\frac{3\pi}{4}\right) = -1$.
$-\sqrt{3} \tan t = -1 \implies \tan t = \frac{1}{\sqrt{3}} \implies t = \frac{\pi}{6}$
At $t = \frac{\pi}{6}$, $x = \sin\left(\frac{\pi}{6}\right) + 1 = 1.5$ and $y = \sqrt{3} \cos\left(\frac{\pi}{6}\right) - 1 = 0.5$.
The gradient of the normal is $-\frac{1}{-1} = 1$, so its equation is:
$y - 0.5 = 1(x - 1.5) \implies y = x - 1$

From (iii), a tangent parallel to the $x$-axis has a gradient of $0$.
$-\sqrt{3} \tan \theta = 0 \implies \theta = 0$
At $\theta = 0$, $y = \sqrt{3} \cos 0 - 1 = \sqrt{3} - 1$.
The equation of this tangent is the horizontal line $y = \sqrt{3} - 1$.

2. Identify the bounded region and set up an integral to calculate its area.

The quadrilateral is bounded by the $y$-axis ($x=0$), the $x$-axis ($y=0$), the normal line ($y=x-1 \implies x=y+1$), and the tangent line ($y=\sqrt{3}-1$).
To avoid splitting the area into multiple sections, we can use horizontal strips and integrate with respect to $y$.
The region is bounded on the right by the line $x=y+1$ and on the left by the $y$-axis ($x=0$), from $y=0$ to $y=\sqrt{3}-1$.
$\text{Area} = \int_{0}^{\sqrt{3}-1} (y + 1) \, \mathrm{d}y$

3. Evaluate the definite integral to find the exact area.

$\text{Area} = \left[ \frac{1}{2}y^2 + y \right]_0^{\sqrt{3}-1}$
$= \frac{1}{2}(\sqrt{3}-1)^2 + (\sqrt{3}-1) - 0$
$= \frac{1}{2}(4 - 2\sqrt{3}) + \sqrt{3} - 1$
$= 2 - \sqrt{3} + \sqrt{3} - 1$
$= 1$