\((a) To solve the equation f(x) = 0, we have:\)

\(x - 4x^{\frac{1}{2}} + 1 = 0\)

\(Let y = x^{\frac{1}{2}}, then x = y^2. Substitute to get:\)

\(y^2 - 4y + 1 = 0\)

\(Using the quadratic formula, y = \frac{4 \pm \sqrt{16 - 4}}{2} = 2 \pm \sqrt{3}\)

\(Thus, x = (2 \pm \sqrt{3})^2 = 7 \pm 4\sqrt{3}\)

(b) Given f(x) \equiv gh(x), we have:

\(gh(x) = m\left(x^{\frac{1}{2}} - 2\right)^2 + n\)

Expanding, we get:

\(gh(x) = m(x - 4x^{\frac{1}{2}} + 4) + n\)

\(Since f(x) = x - 4x^{\frac{1}{2}} + 1, equate the expressions:\)

\(m(x - 4x^{\frac{1}{2}} + 4) + n = x - 4x^{\frac{1}{2}} + 1\)

\(Comparing coefficients, m = 1 and n = -3.\)