(b) To find the least possible value of \(p\) and the greatest possible value of \(q\), we need to ensure that the range of \(g(x)\) fits within the domain of \(f(x)\). The function \(g(x) = 3x + 1\) is defined for \(x < 8\), so \(g(x) < 25\). We need \(p < g(x) < q\).

Solving \(4x^2 - 12x + 13 < 8\), we get:

\((2x - 3)^2 + 4 < 8\)

\((2x - 3)^2 < 4\)

\(-2 < 2x - 3 < 2\)

\(\frac{1}{2} < x < 2\frac{1}{2}\)

Thus, the least \(p = \frac{1}{2}\) and the greatest \(q = 2\frac{1}{2}\).

(c) To find \(gf(x)\), substitute \(g(x) = 3x + 1\) into \(f(x)\):

\(gf(x) = f(g(x)) = f(3x + 1)\)

\(gf(x) = 4(3x + 1)^2 - 12(3x + 1) + 13\)

\(gf(x) = 4(9x^2 + 6x + 1) - 36x - 12 + 13\)

\(gf(x) = 36x^2 + 24x + 4 - 36x - 12 + 13\)

\(gf(x) = 12x^2 - 36x + 40\)