(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, \(y = -\left( \frac{3}{2} \cos 2x + \frac{1}{2} \right) = -\frac{3}{2} \cos 2x - \frac{1}{2}\).