(i) The gradient of the normal line \(3y + x = 17\) is \(-\frac{1}{3}\). Since the line is normal to the curve, the product of the gradients of the tangent and the normal is \(-1\). Therefore, the gradient of the tangent \(\frac{dy}{dx} = 3\).

Setting \(5 - \frac{8}{x^2} = 3\), we solve for \(x\):

\(5 - \frac{8}{x^2} = 3\)

\(\frac{8}{x^2} = 2\)

\(x^2 = 4\)

\(x = 2\) (since \(x\) is positive)

Substitute \(x = 2\) into the normal line equation to find \(y\):

\(3y + 2 = 17\)

\(3y = 15\)

\(y = 5\)

Thus, the coordinates of \(P\) are \((2, 5)\).

(ii) To find the equation of the curve, integrate \(\frac{dy}{dx} = 5 - \frac{8}{x^2}\):

\(y = \int (5 - \frac{8}{x^2}) \, dx\)

\(y = 5x + 8x^{-1} + c\)

Using the point \((2, 5)\) to find \(c\):

\(5 = 5(2) + \frac{8}{2} + c\)

\(5 = 10 + 4 + c\)

\(c = -9\)

Therefore, the equation of the curve is \(y = 5x + \frac{8}{x} - 9\).