Answer: \(2\lt m\lt 26\).

At an intersection,

\(mx-2=(m+1)x^2+8x+1.\)

Rearrange into a quadratic in \(x\):

\((m+1)x^2+(8-m)x+3=0.\)

For the line not to touch or cut the curve, this quadratic must have no real roots. Therefore its discriminant must be negative:

\((8-m)^2-4(m+1)(3)\lt 0.\)

Simplify:

\((8-m)^2-12(m+1)\lt 0,\)

\(m^2-16m+64-12m-12\lt 0,\)

\(m^2-28m+52\lt 0.\)

Factorise:

\((m-2)(m-26)\lt 0.\)

The product is negative between the roots, so

\(2\lt m\lt 26.\)