Answer: \(k\lt \frac{2}{9}\) or \(k\gt 2\).

For a quadratic equation to have real and distinct roots, its discriminant must be positive.

The equation is

\(2x^2+(3k-2)x+k=0.\)

Here

\(a=2,\quad b=3k-2,\quad c=k.\)

So

\((3k-2)^2-4(2)(k)\gt 0.\)

Expand and simplify:

\(9k^2-12k+4-8k\gt 0.\)

Therefore

\(9k^2-20k+4\gt 0.\)

Factorise:

\((9k-2)(k-2)\gt 0.\)

The critical values are

\(k=\frac29\quad\text{and}\quad k=2.\)

The quadratic expression in \(k\) has positive leading coefficient, so it is positive outside the two roots.

Hence the possible values of \(k\) are

\(k\lt \frac29\quad\text{or}\quad k\gt 2.\)