(i) To find the area under the curve \(y = x^2 e^{-x}\) from \(x = 0\) to \(x = 3\), we integrate:

\(\int_0^3 x^2 e^{-x} \, dx\)

Using integration by parts, let \(u = x^2\) and \(dv = e^{-x} \, dx\). Then \(du = 2x \, dx\) and \(v = -e^{-x}\).

\(\int x^2 e^{-x} \, dx = -x^2 e^{-x} + \int 2x e^{-x} \, dx\)

Apply integration by parts again to \(\int 2x e^{-x} \, dx\):

Let \(u = 2x\) and \(dv = e^{-x} \, dx\). Then \(du = 2 \, dx\) and \(v = -e^{-x}\).

\(\int 2x e^{-x} \, dx = -2x e^{-x} + \int 2 e^{-x} \, dx\)

\(= -2x e^{-x} - 2e^{-x}\)

Substitute back:

\(\int x^2 e^{-x} \, dx = -x^2 e^{-x} - 2x e^{-x} - 2e^{-x}\)

Evaluate from \(x = 0\) to \(x = 3\):

\(\left[ -x^2 e^{-x} - 2x e^{-x} - 2e^{-x} \right]_0^3\)

At \(x = 3\):

\(-9e^{-3} - 6e^{-3} - 2e^{-3} = -17e^{-3}\)

At \(x = 0\):

\(0 - 0 - 2 = -2\)

Thus, the area is:

\(-(-2) - (-17e^{-3}) = 2 - \frac{17}{e^3}\)

(ii) To find the maximum point, differentiate \(y = x^2 e^{-x}\):

\(\frac{dy}{dx} = 2x e^{-x} - x^2 e^{-x} = e^{-x}(2x - x^2)\)

Set \(\frac{dy}{dx} = 0\):

\(e^{-x}(2x - x^2) = 0\)

\(2x - x^2 = 0\)

\(x(2 - x) = 0\)

\(x = 0\) or \(x = 2\). Since \(x = 0\) is not a maximum, \(x = 2\) is the maximum point.

(iii) The tangent at \(P\) passes through the origin. The equation of the tangent at \(x = a\) is:

\(y - a^2 e^{-a} = (2a - a^2)e^{-a}(x - a)\)

Substitute \((0,0)\):

\(0 - a^2 e^{-a} = (2a - a^2)e^{-a}(0 - a)\)

\(a^2 = a^2(2 - a)\)

\(2 - a = 1\)

\(a = 1\). Thus, \(x = 1\).