Answer: \(k\lt \frac19\) or \(k\gt 1\).

At an intersection point,

\(x-3=k^2x^2+5kx+1.\)

Bring all terms to one side:

\(k^2x^2+(5k-1)x+4=0.\)

For two distinct intersections, this quadratic in \(x\) must have two distinct roots, so its discriminant must be positive:

\((5k-1)^2-4(k^2)(4)\gt 0.\)

Hence

\(25k^2-10k+1-16k^2\gt 0,\)

so

\(9k^2-10k+1\gt 0.\)

Factorise:

\((9k-1)(k-1)\gt 0.\)

The critical values are \(k=\frac19\) and \(k=1\). Since the quadratic in \(k\) is positive outside these roots,

\(\boxed{k\lt \frac19\text{ or }k\gt 1}.\)