(a) To find the equation of the tangent, first differentiate the curve equation \(y = x^{\frac{1}{2}} + 4x^{-\frac{1}{2}}\).

The derivative is \(\frac{dy}{dx} = \frac{1}{2}x^{-\frac{1}{2}} - 2x^{-\frac{3}{2}}\).

At \(x = 1\), \(\frac{dy}{dx} = \frac{1}{2} \times 1^{-\frac{1}{2}} - 2 \times 1^{-\frac{3}{2}} = \frac{1}{2} - 2 = -\frac{3}{2}\).

The equation of the tangent at \(A(1, 5)\) is \(y - 5 = -\frac{3}{2}(x - 1)\).

Simplifying gives \(y = -\frac{3}{2}x + \frac{13}{2}\).

(b) To find the area of the shaded region, integrate the curve equation from \(x = 1\) to \(x = 16\).

The integral of \(y = x^{\frac{1}{2}} + 4x^{-\frac{1}{2}}\) is \(\frac{x^{\frac{3}{2}}}{\frac{3}{2}} + 8x^{\frac{1}{2}}\).

Evaluating from 1 to 16 gives \(\left[ \frac{128}{3} + 32 \right] - \left[ \frac{2}{3} + 8 \right] = 66\).

The area under the line \(y = 5\) from \(x = 1\) to \(x = 16\) is \(15 \times 5 = 75\).

The required area is \(75 - 66 = 9\).