Answer: Intercepts \(\left(-\frac12,0\right)\), \((5,0)\), \((0,5)\); exactly two solutions when \(k=0\) or \(k\gt \frac{121}{8}\).

The graph meets the \(x\)-axis when

\(2x^2-9x-5=0.\)

Factorising gives

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

So the \(x\)-intercepts are

\(\left(-\frac12,0\right)\quad\text{and}\quad(5,0).\)

The \(y\)-intercept is

\(\left|2(0)^2-9(0)-5\right|=5,\)

so the graph meets the y-axis at \((0,5)\).

The original quadratic \(2x^2-9x-5\) is negative between the roots, so that part is reflected above the \(x\)-axis by the modulus.

The maximum of this reflected middle section occurs at

\(x=\frac{-(-9)}{2(2)}=\frac94.\)

At this value,

\(2\left(\frac94\right)^2-9\left(\frac94\right)-5=-\frac{121}{8}.\)

So the reflected maximum is

\(\frac{121}{8}.\)

For \(\left|2x^2-9x-5\right|=k\) to have exactly two solutions, either the horizontal line is the \(x\)-axis, giving the two roots, or it lies above the reflected maximum.

Therefore

\(\boxed{k=0\quad\text{or}\quad k\gt \frac{121}{8}}.\)