曲线 $C$ 的参数方程为
$x = 1 - 2\cos 2\theta, \quad y = \sin 2\theta - 1, \quad \text{其中 } 0 \le \theta \le \frac{3\pi}{8}.$

1. Set $x = 0$ to find the value of the parameter $\theta$ at the $y$-intercept.

At the point where the curve cuts the $y$-axis, $x = 0$:
$1 - 2\cos 2\theta = 0$
$\cos 2\theta = \frac{1}{2}$
Given the domain $0 \le \theta \le \frac{3\pi}{8}$, the range for $2\theta$ is $0 \le 2\theta \le \frac{3\pi}{4}$.
The only solution in this interval is:
$2\theta = \frac{\pi}{3}$
$\theta = \frac{\pi}{6}$

2. Substitute the value of $2\theta$ into the parametric equation for $y$ to determine the exact coordinates.

Substitute $2\theta = \frac{\pi}{3}$ into the equation for $y$:
$y = \sin\left(\frac{\pi}{3}\right) - 1$
$y = \frac{\sqrt{3}}{2} - 1$
The exact coordinates of the point are $\left(0, \frac{\sqrt{3}}{2} - 1\right)$.

1. Differentiate $x$ and $y$ with respect to $\theta$.

$\frac{\mathrm{d}x}{\mathrm{d}\theta} = -2(-2\sin 2\theta) = 4\sin 2\theta$
$\frac{\mathrm{d}y}{\mathrm{d}\theta} = 2\cos 2\theta$

2. Apply the chain rule to find $\frac{\mathrm{d}y}{\mathrm{d}x}$.

$\frac{\mathrm{d}y}{\mathrm{d}x} = \frac{\frac{\mathrm{d}y}{\mathrm{d}\theta}}{\frac{\mathrm{d}x}{\mathrm{d}\theta}}$
$\frac{\mathrm{d}y}{\mathrm{d}x} = \frac{2\cos 2\theta}{4\sin 2\theta}$
$\frac{\mathrm{d}y}{\mathrm{d}x} = \frac{1}{2}\cot 2\theta$ (shown)

3. Find the coordinates at $\theta = \frac{\pi}{4}$.

When $\theta = \frac{\pi}{4}$:
$x = 1 - 2\cos\left(2 \times \frac{\pi}{4}\right) = 1 - 2\cos\left(\frac{\pi}{2}\right) = 1 - 0 = 1$
$y = \sin\left(2 \times \frac{\pi}{4}\right) - 1 = \sin\left(\frac{\pi}{2}\right) - 1 = 1 - 1 = 0$
The coordinates of the point are $(1, 0)$.

4. Find the gradient at $\theta = \frac{\pi}{4}$.

When $\theta = \frac{\pi}{4}$:
$\frac{\mathrm{d}y}{\mathrm{d}x} = \frac{1}{2}\cot\left(2 \times \frac{\pi}{4}\right) = \frac{1}{2}\cot\left(\frac{\pi}{2}\right) = \frac{1}{2}(0) = 0$
The gradient is $0$.

5. Analyse the behaviour of the tangent as $\theta \to 0$.

As $\theta \to 0$, $2\theta \to 0^+$, so $\cot 2\theta \to \infty$.
Therefore, $\frac{\mathrm{d}y}{\mathrm{d}x} \to \infty$.
This means the tangent to $C$ is vertical (parallel to the $y$-axis).

1. Calculate the exact coordinates of the endpoints by substituting the boundary values of $\theta$.

For $\theta = 0$:
$x = 1 - 2\cos(0) = -1$
$y = \sin(0) - 1 = -1$
Endpoint 1: $(-1, -1)$

For $\theta = \frac{3\pi}{8}$:
$x = 1 - 2\cos\left(\frac{3\pi}{4}\right) = 1 - 2\left(-\frac{\sqrt{2}}{2}\right) = 1 + \sqrt{2}$
$y = \sin\left(\frac{3\pi}{4}\right) - 1 = \frac{\sqrt{2}}{2} - 1$
Endpoint 2: $\left(1+\sqrt{2}, \frac{\sqrt{2}}{2}-1\right)$

2. Identify the axial intercepts and summarize the key tangent features from previous parts.

From part (a), the $y$-intercept is $\left(0, \frac{\sqrt{3}}{2}-1\right)$.
From part (b), when $\theta = \frac{\pi}{4}$, the point is $(1, 0)$, which serves as the $x$-intercept. The gradient $\frac{\mathrm{d}y}{\mathrm{d}x} = 0$, indicating a horizontal tangent.
Also from part (b), as $\theta \to 0$, $\frac{\mathrm{d}y}{\mathrm{d}x} \to \infty$, so the tangent to $C$ at $(-1, -1)$ is vertical.

3. Construct the final sketch incorporating all evaluated points and features.

Draw a smooth, continuous curve matching the gathered features:

1. Start at $(-1, -1)$, drawing the curve initially tangent to the vertical line $x = -1$.
2. Curve upwards to cross the $y$-axis precisely at $\left(0, \frac{\sqrt{3}}{2}-1\right)$.
3. Reach a maximum at $(1, 0)$, showing the curve is perfectly horizontal (tangent to the $x$-axis) at this point.
4. Continue downwards and to the right, terminating at the exact point $\left(1+\sqrt{2}, \frac{\sqrt{2}}{2}-1\right)$.
   All coordinates must be explicitly labelled on the sketch.

1. Rearrange the parametric equations to make $\cos 2\theta$ and $\sin 2\theta$ the subjects.

From $x = 1 - 2\cos 2\theta$, we get:
$\cos 2\theta = \frac{1-x}{2}$

From $y = \sin 2\theta - 1$, we get:
$\sin 2\theta = y+1$

2. Use the Pythagorean identity to eliminate the parameter $\theta$.

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

3. Determine the domain of the cartesian equation based on the restricted values of $\theta$.

Given the restriction $0 \le \theta \le \frac{3\pi}{8}$, we have:
$0 \le 2\theta \le \frac{3\pi}{4}$

The range of $\cos 2\theta$ on this interval is $\left[-\frac{\sqrt{2}}{2}, 1\right]$.
Substituting these extremes into $x = 1 - 2\cos 2\theta$:
Minimum value of $x = 1 - 2(1) = -1$
Maximum value of $x = 1 - 2\left(-\frac{\sqrt{2}}{2}\right) = 1 + \sqrt{2}$

Thus, the domain is $-1 \le x \le 1 + \sqrt{2}$.