Answer: \(k=4\) or \(k=12\).

At a point of intersection,

\(kx+3=2x^2+4x+k-1.\)

Rearrange this as a quadratic equation in \(x\):

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

For the line to be a tangent to the curve, this quadratic must have exactly one repeated root. Therefore its discriminant is zero:

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

So

\(k^2-8k+16-8k+32=0.\)

Hence

\(k^2-16k+48=0.\)

Factorising,

\((k-4)(k-12)=0.\)

Therefore

\(k=4\quad\text{or}\quad k=12.\)