(a) To find the equation of the tangent, we first find the gradient at \(x = 6\) using \(\frac{dy}{dx} = 12\left(\frac{1}{2}x - 1\right)^{-4}\).

Substitute \(x = 6\):

\(12\left(\frac{1}{2} \times 6 - 1\right)^{-4} = 12(2)^{-4} = \frac{3}{4}\)

The gradient of the tangent is \(\frac{3}{4}\). Using the point-slope form \(y - y_1 = m(x - x_1)\) with \(P(6, 4)\):

\(y - 4 = \frac{3}{4}(x - 6)\)

Simplifying gives:

\(y = \frac{3}{4}x - \frac{1}{2}\)

(b) To find the equation of the curve, integrate \(\frac{dy}{dx} = 12\left(\frac{1}{2}x - 1\right)^{-4}\).

\(y = \int 12\left(\frac{1}{2}x - 1\right)^{-4} \, dx\)

Let \(u = \frac{1}{2}x - 1\), then \(du = \frac{1}{2}dx\) or \(dx = 2du\).

\(y = \int 12u^{-4} \, 2du = 24\int u^{-4} \, du\)

\(y = 24\left(\frac{u^{-3}}{-3}\right) + c = -8u^{-3} + c\)

Substitute back \(u = \frac{1}{2}x - 1\):

\(y = -8\left(\frac{1}{2}x - 1\right)^{-3} + c\)

Using point \(P(6, 4)\) to find \(c\):

\(4 = -8(2)^{-3} + c\)

\(4 = -1 + c\)

\(c = 5\)

Thus, the equation of the curve is:

\(y = -8\left(\frac{1}{2}x - 1\right)^{-3} + 5\)