(a) Substitute \(y = \frac{1}{2}x + 6\) into the circle's equation:

\((x-a)^2 + \left(\frac{1}{2}x + 6 - 3\right)^2 = 20\)

\((x-a)^2 + \left(\frac{1}{2}x + 3\right)^2 = 20\)

\(\frac{5}{4}x^2 + (3-2a)x + a^2 - 11 = 0\)

Using the quadratic formula, solve for \(a\):

\((3-2a)^2 - 4\cdot\frac{5}{4}(a^2-1) = 0\)

\(a = 4\) or \(a = -16\)

(b) For \(a = 4\), the center of the circle is \((4, 3)\) and \(P = (2, 7)\). The gradient of the normal is \(-2\).

The equation of the normal is:

\(y - 3 = -2(x - 4)\) or \(y - 7 = -2(x - 2)\)

(c) The tangents parallel to the normal have the same gradient \(-2\). Using the circle's equation:

\((x-4)^2 + \left(\frac{1}{2}x + 3\right)^2 = 20\)

\(\frac{5}{4}x^2 - 10x = 0\)

\(x = 0\) or \(x = 8\)

Coordinates are \((0, 1)\) and \((8, 5)\).

Equations of tangents:

\(x + y = 1\) and \(2x + y = 21\)