The transformation applied to the lower curve \(y = \\cos \\theta\) to obtain the upper curve involves both a vertical stretch and a horizontal stretch.

The vertical stretch is by a factor of 2, and the horizontal stretch is by a factor of 2, which affects the period of the cosine function. Additionally, there is a vertical translation upwards by 3 units.

Thus, the equation of the upper curve is:

\(y = 3 + 2 \cos \left( \frac{1}{2} \theta \right)\)

(a) The function \(f(x) = 2 - 3 \sin 2x\) has a range determined by the values of \(\sin 2x\), which varies from \(-1\) to \(1\). Therefore, \(f(x)\) ranges from \(2 - 3(1) = -1\) to \(2 - 3(-1) = 5\).

The function \(g(x) = -2f(x)\) will have a range that is twice the range of \(f(x)\) and inverted. Thus, \(g(x)\) ranges from \(-2(5) = -10\) to \(-2(-1) = 2\).

(b) The graph of \(y = g(x)\) is a reflection of \(y = f(x)\) in the x-axis and stretched by a factor of 2 in the y direction.

(c) To map \(y = f(x)\) onto \(y = h(x)\), perform the following transformations:

• Reflect in the x-axis.
• Stretch by a factor of 2 in the y direction.
• Translate by \(-\pi\) in the x direction, or equivalently, translate by \(\begin{pmatrix} 0 \\ -\pi \end{pmatrix}\).

(a) The function \(f(x) = \frac{3}{2} \cos 2x + \frac{1}{2}\) has a cosine component \(\frac{3}{2} \cos 2x\) which varies between \(-\frac{3}{2}\) and \(\frac{3}{2}\). Adding \(\frac{1}{2}\) shifts this range to \(-1\) to \(2\). Thus, the range of \(f(x)\) is \(-1 < f(x) < 2\).

(b) Since the x-axis is a tangent to \(y = g(x)\), the minimum value of \(g(x)\) must be 0. Therefore, \(f(x) + k = 0\) when \(f(x)\) is at its minimum, which is \(-1\). Solving \(-1 + k = 0\) gives \(k = 1\). The transformation is a translation by 1 unit upwards parallel to the y-axis.

(c) The reflection of \(y = f(x)\) in the x-axis is given by \(y = -f(x)\). Therefore, \(y = -\left( \frac{3}{2} \cos 2x + \frac{1}{2} \right) = -\frac{3}{2} \cos 2x - \frac{1}{2}\).