Answer: (i) \(k=6\) or \(k=10\). (ii) \((2,1)\).

(i) At a point of contact, the tangent and curve have exactly one common point. Equate the line and curve:

\(2k+1-kx=12-4x-x^2.\)

Bring all terms to one side:

\(x^2+(4-k)x+(2k-11)=0.\)

For tangency, this quadratic has a repeated root, so its discriminant is zero:

\((4-k)^2-4(2k-11)=0.\)

Expand and simplify:

\(k^2-8k+16-8k+44=0,\)

so

\(k^2-16k+60=0.\)

Factorise:

\((k-6)(k-10)=0.\)

Thus

\(k=6\quad\text{or}\quad k=10.\)

(ii) If \(k=6\), the tangent is

\(y=13-6x.\)

If \(k=10\), the tangent is

\(y=21-10x.\)

At their intersection,

\(13-6x=21-10x.\)

So

\(4x=8, \qquad x=2.\)

Substitute into \(y=13-6x\):

\(y=13-12=1.\)

Therefore the point of intersection is

\(\boxed{(2,1)}.\)