(i) Differentiate \(x = a \cos^4 t\) and \(y = a \sin^4 t\) with respect to \(t\):

\(\frac{dx}{dt} = -4a \cos^3 t \sin t\)

\(\frac{dy}{dt} = 4a \sin^3 t \cos t\)

Then, \(\frac{dy}{dx} = \frac{\frac{dy}{dt}}{\frac{dx}{dt}} = \frac{4a \sin^3 t \cos t}{-4a \cos^3 t \sin t} = -\tan t\).

(ii) The equation of the tangent is given by:

\(\frac{y - a \sin^4 t}{x - a \cos^4 t} = -\tan t\)

Rearranging gives:

\(x \sin^2 t + y \cos^2 t = a \sin^2 t \cos^2 t\).

(iii) For the tangent meeting the x-axis at \(P\) and the y-axis at \(Q\):

At \(P\), \(y = 0\), so \(x = a \cos^2 t\).

At \(Q\), \(x = 0\), so \(y = a \sin^2 t\).

Thus, \(OP = a \cos^2 t\) and \(OQ = a \sin^2 t\).

Therefore, \(OP + OQ = a \cos^2 t + a \sin^2 t = a(\cos^2 t + \sin^2 t) = a\).