曲线 C 由以下参数方程定义： $x = \cos t + \dfrac{1}{2} \cos 5t, \quad y = \sin t + \dfrac{1}{2} \sin 5t \quad$（其中 $0 \le t \le \dfrac{\pi}{2}$）。 另一条曲线 D 具有以下参数方程： $x = 2 + \sqrt{k} \cos \theta, \quad y = \dfrac{3}{2} \sin \theta \quad$（其中 $0 \le \theta \le 2\pi$ 且 $k > 0$）。

1. Substitute $t = 0$ into the parametric equations of curve C to find the corresponding $x$ and $y$ values.

Substitute $t = 0$ into the equation for $x$:
$x = \cos(0) + \frac{1}{2} \cos(0)$
$x = 1 + \frac{1}{2}(1) = \frac{3}{2}$

Substitute $t = 0$ into the equation for $y$:
$y = \sin(0) + \frac{1}{2} \sin(0)$
$y = 0 + \frac{1}{2}(0) = 0$

The coordinates are $\left( \frac{3}{2}, 0 \right)$.

1. Express $\cos \theta$ and $\sin \theta$ in terms of $x$ and $y$ respectively.

$x = 2 + \sqrt{k} \cos \theta \implies \cos \theta = \frac{x - 2}{\sqrt{k}}$

$y = \frac{3}{2} \sin \theta \implies \sin \theta = \frac{2y}{3}$

2. Use the Pythagorean identity to eliminate the parameter $\theta$ and obtain the Cartesian equation.

Using the identity $\cos^2 \theta + \sin^2 \theta = 1$:

$\left( \frac{x - 2}{\sqrt{k}} \right)^2 + \left( \frac{2y}{3} \right)^2 = 1$

$\frac{(x - 2)^2}{k} + \frac{4y^2}{9} = 1$

1. Sketch curve C by interpreting the given parametric equations using a graphing calculator.

Using a graphing calculator, plot the parametric equations for curve C: $x = \cos t + \frac{1}{2}\cos 5t$, $y = \sin t + \frac{1}{2}\sin 5t$ for $0 \le t \le \frac{\pi}{2}$.
The calculator will display a self-intersecting (looped) curve strictly within the first quadrant. The starting point at $t=0$ is $(1.5, 0)$ and the ending point at $t=\frac{\pi}{2}$ is $(0, 1.5)$. Mark these coordinates on the axes.

2. Sketch curve D using the standard properties of its cartesian equation.

From part (b), the cartesian equation for curve D is $\frac{(x-2)^2}{k} + \frac{y^2}{9/4} = 1$.
This represents an ellipse centered at $(2, 0)$.
The vertical semi-axis length is $\sqrt{\frac{9}{4}} = 1.5$, giving the highest point $(2, 1.5)$ and lowest point $(2, -1.5)$. Note that the highest point aligns horizontally with the endpoint $(0, 1.5)$ of curve C.
The horizontal semi-axis length is $\sqrt{k}$, giving the $x$-intercepts $(2-\sqrt{k}, 0)$ and $(2+\sqrt{k}, 0)$.
Draw a representative full ellipse for an arbitrary $k > 0$, labelling these key points.

1. Link the given equation to the intersection of Curve C and Curve D.

Substitute the parametric equations of Curve C, $x = \cos t + 0.5\cos 5t$ and $y = \sin t + 0.5\sin 5t$, into the given equation:
$\dfrac{\left(x - 2\right)^2}{k} + \dfrac{y^2}{9/4} = 1$
This matches the cartesian equation of Curve D found in part (b). Thus, the given equation represents points where Curve C intersects Curve D. For the equation to have no real solutions, the two curves must not intersect.

2. Determine the maximum $x$-coordinate of Curve C.

From part (a), the starting point of Curve C (at $t = 0$) is $(1.5, 0)$.
Since $\cos t \le 1$ and $\cos 5t \le 1$, the maximum possible value for $x$ on Curve C is:
$x_{max} = 1 + 0.5(1) = 1.5$
Thus, Curve C lies entirely in the region $x \le 1.5$, connecting $(1.5, 0)$ to $(0, 1.5)$.

3. Determine the minimum $x$-coordinate of Curve D.

Curve D is an ellipse centered at $(2, 0)$ with a horizontal semi-axis of $\sqrt{k}$ and a vertical semi-axis of $\frac{3}{2}$.
The left-most point (minimum $x$-coordinate) of the ellipse is $x_{min} = 2 - \sqrt{k}$.

4. Establish the geometric condition for no intersection and solve for $k$.

For the curves to not intersect, Curve D must lie strictly to the right of Curve C. We require the minimum $x$ of Curve D to be strictly greater than the maximum $x$ of Curve C:
$2 - \sqrt{k} > 1.5$
$\sqrt{k} < 0.5$
$k < 0.25$
(Note: If $k \ge 0.25$, the point $(1.5, 0)$ of Curve C lies inside or on the ellipse. Since the highest point of Curve C, $(0, 1.5)$, is always outside the ellipse — as $D$'s maximum $y$ is $1.5$ occurring exclusively at $x=2$ — Curve C will inevitably cross Curve D, guaranteeing real solutions.)

Given the condition $k > 0$, the range of values is:
$0 < k < 0.25$