Answer: \(k=-\dfrac{19}{8}\), and \(D=\left(\dfrac{29}{8},-\dfrac98\right)\).

We start with the main method. Substitute the line into the curve; a tangent corresponds to a repeated root, so the discriminant is zero.

Substitute \(y=3x-2\) into

\(2x^2-xy+y^2=2.\)

This gives

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

Expanding and simplifying,

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

Factorise:

\((4x-1)(x-1)=0.\)

So

\(x=\frac14\quad\text{or}\quad x=1.\)

Using \(y=3x-2\), the intersection points are

\(\left(\frac14,-\frac54\right)\quad\text{and}\quad(1,1).\)

The midpoint of \(AB\) is

\(\left(\frac{\frac14+1}{2},\frac{-\frac54+1}{2}\right) =\left(\frac58,-\frac18\right).\)

The gradient of \(AB\) is \(3\), because \(AB\) lies on \(y=3x-2\).

Therefore the perpendicular bisector has gradient \(-\frac13\).

Its equation is

\(y+\frac18=-\frac13\left(x-\frac58\right).\)

The point \(C=\left(k,\frac78\right)\) lies on this line, so substitute \(y=\frac78\):

\(\frac78+\frac18=-\frac13\left(k-\frac58\right).\)

So

\(1=-\frac13\left(k-\frac58\right).\)

Hence

\(k-\frac58=-3\), and

\(k=-\frac{19}{8}.\)

For part (b), \(D\) is the reflection of \(C\) in the line \(AB\).

The perpendicular bisector meets \(AB\) at the midpoint of \(CD\), which is also the midpoint of \(AB\):

\(M=\left(\frac58,-\frac18\right).\)

Since \(M\) is the midpoint of \(C\) and \(D\),

\(D=2M-C.\)

Thus

\(D=\left(2\cdot\frac58-\left(-\frac{19}{8}\right),\ 2\cdot\left(-\frac18\right)-\frac78\right).\)

Therefore

\(D=\left(\frac{29}{8},-\frac98\right).\)

This completes the solution and gives the required result.