To find the values of \(c\) and coordinates of \(P\), we set the equations equal since the line is tangent to the curve:

\(cx^2 + 2x - 3 = 6x - c\)

Rearrange to form a quadratic equation:

\(cx^2 + 2x - 3 - 6x + c = 0\)

\(cx^2 - 4x + (c - 3) = 0\)

For the line to be tangent, the discriminant must be zero:

\(b^2 - 4ac = 0\)

Here, \(a = c\), \(b = -4\), \(c = c - 3\).

\((-4)^2 - 4(c)(c - 3) = 0\)

\(16 - 4c^2 + 12c = 0\)

\(4c^2 - 12c - 16 = 0\)

Divide by 4:

\(c^2 - 3c - 4 = 0\)

Factorize:

\((c - 4)(c + 1) = 0\)

Thus, \(c = 4\) or \(c = -1\).

For \(c = 4\):

Substitute \(c = 4\) into the quadratic equation:

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

\((2x - 1)^2 = 0\)

\(x = \frac{1}{2}\)

Substitute \(x = \frac{1}{2}\) into \(y = 6x - c\):

\(y = 6 \times \frac{1}{2} - 4 = -1\)

Coordinates of \(P\) are \(\left( \frac{1}{2}, -1 \right)\).

For \(c = -1\):

Substitute \(c = -1\) into the quadratic equation:

\(x^2 + 4x + 4 = 0\)

\((x + 2)^2 = 0\)

\(x = -2\)

Substitute \(x = -2\) into \(y = 6x - c\):

\(y = 6 \times (-2) + 1 = -11\)

Coordinates of \(P\) are \((-2, -11)\).

To find the points of intersection, substitute the expression for x from the line equation into the curve equation.

From x + 2y = 9, we have x = 9 - 2y.

Substitute into the curve equation xy + 18 = 0:

\((9 - 2y)y + 18 = 0\)

\(9y - 2y^2 + 18 = 0\)

\(-2y^2 + 9y + 18 = 0\)

Rearrange to form a quadratic equation:

\(2y^2 - 9y - 18 = 0\)

Solving this quadratic equation using the quadratic formula:

\(y = \frac{-b \pm \sqrt{b^2 - 4ac}}{2a}\)

where \(a = 2\), \(b = -9\), \(c = -18\).

Calculate the discriminant:

\(b^2 - 4ac = (-9)^2 - 4 \times 2 \times (-18) = 81 + 144 = 225\)

Thus,

\(y = \frac{9 \pm 15}{4}\)

So, \(y = 6\) or \(y = -1.5\).

Substitute back to find \(x\):

For \(y = 6\):

\(x = 9 - 2(6) = -3\)

For \(y = -1.5\):

\(x = 9 - 2(-1.5) = 12\)

Thus, the coordinates of the points are (12, -1.5) and (-3, 6).

To find the values of \(k\) and \(a\), we equate the equations of the curve and the line at the given \(x\)-coordinates.

1. For \(x = 0\):

\(4(0)^2 - k(0) + \frac{1}{2}k^2 = 0 - a\)

\(\frac{1}{2}k^2 = -a\) (Equation 1)

2. For \(x = \frac{3}{4}\):

\(4\left(\frac{3}{4}\right)^2 - k\left(\frac{3}{4}\right) + \frac{1}{2}k^2 = \frac{3}{4} - a\)

\(4\left(\frac{9}{16}\right) - \frac{3}{4}k + \frac{1}{2}k^2 = \frac{3}{4} - a\)

\(\frac{9}{4} - \frac{3}{4}k + \frac{1}{2}k^2 = \frac{3}{4} - a\) (Equation 2)

Substitute \(a = -\frac{1}{2}k^2\) from Equation 1 into Equation 2:

\(\frac{9}{4} - \frac{3}{4}k + \frac{1}{2}k^2 = \frac{3}{4} + \frac{1}{2}k^2\)

\(\frac{9}{4} - \frac{3}{4}k = \frac{3}{4}\)

\(\frac{9}{4} - \frac{3}{4} = \frac{3}{4}k\)

\(\frac{6}{4} = \frac{3}{4}k\)

\(k = 2\)

Substitute \(k = 2\) back into Equation 1:

\(\frac{1}{2}(2)^2 = -a\)

\(2 = -a\)

\(a = -2\)

Thus, \(k = 2\) and \(a = -2\).

To find the values of \(m\), equate the line and curve equations: \(mx - 6 = x^2 - 4x + 3\).

Rearrange to form a quadratic: \(x^2 - (4 + m)x + 9 = 0\).

For the line to be tangent, the discriminant must be zero: \(b^2 - 4ac = 0\).

Here, \(a = 1\), \(b = -(4 + m)\), \(c = 9\).

So, \((4 + m)^2 - 4 \times 1 \times 9 = 0\).

\((4 + m)^2 = 36\).

\(4 + m = 6\) or \(4 + m = -6\).

\(m = 2\) or \(m = -10\).

For \(m = 2\), substitute back to find \(x\):

\(x^2 - 6x + 9 = 0\) gives \((x - 3)^2 = 0\), so \(x = 3\).

Substitute \(x = 3\) into \(y = mx - 6\):

\(y = 2 \times 3 - 6 = 0\).

Point is \((3, 0)\).

For \(m = -10\), substitute back to find \(x\):

\(x^2 + 6x + 9 = 0\) gives \((x + 3)^2 = 0\), so \(x = -3\).

Substitute \(x = -3\) into \(y = mx - 6\):

\(y = -10 \times (-3) - 6 = 24\).

Point is \((-3, 24)\).

To find the \(x\)-coordinates of the intersection points, set the equations equal: \(7\sqrt{x} = 6x + 2\).

Rearrange to form a quadratic equation: \(6x + 2 = 7\sqrt{x} \Rightarrow 6(\sqrt{x})^2 - 7\sqrt{x} + 2 = 0\).

Let \(t = \sqrt{x}\), then the equation becomes \(6t^2 - 7t + 2 = 0\).

Factor the quadratic: \((3t - 2)(2t - 1) = 0\).

Solving for \(t\), we get \(t = \frac{2}{3}\) or \(t = \frac{1}{2}\).

Since \(t = \sqrt{x}\), we have \(\sqrt{x} = \frac{2}{3}\) or \(\sqrt{x} = \frac{1}{2}\).

Thus, \(x = \left(\frac{2}{3}\right)^2 = \frac{4}{9}\) or \(x = \left(\frac{1}{2}\right)^2 = \frac{1}{4}\).