Answer: (i) other root \(4\), \(a=-10\), \(b=33\). (ii) \(6,6,1\); \(2,2,9\); \(1,1,36\).

For a monic cubic \(x^3+ax^2+bx-36=0\), the product of the three roots is \(36\).

If \(x=3\) is the repeated root, the roots are \(3\), \(3\), and another root \(r\).

So

\(3\cdot3\cdot r=36.\)

Hence

\(r=4.\)

The cubic is therefore

\((x-3)^2(x-4).\)

Expand:

\((x^2-6x+9)(x-4)=x^3-10x^2+33x-36.\)

Thus

\(a=-10,\qquad b=33.\)

For the other possibilities, let the repeated positive integer root be \(r\) and the third integer root be \(s\).

Then

\(r^2s=36.\)

The possible repeated positive integer roots, excluding \(r=3\), are \(r=1\), \(r=2\), and \(r=6\).

If \(r=1\), then \(s=36\), so the roots are \(1,1,36\).

If \(r=2\), then \(s=9\), so the roots are \(2,2,9\).

If \(r=6\), then \(s=1\), so the roots are \(6,6,1\).

Therefore the other possible sets of roots are

\(\boxed{6,6,1;\quad 2,2,9;\quad 1,1,36}.\)