To find the value of \(m\) for which the line is tangent to the curve, we set the equations equal: \(x^2 - 4x + 4 = mx\).

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

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

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

Calculate the discriminant: \((4 + m)^2 - 4(1)(4) = 0\).

Simplify: \((4 + m)^2 - 16 = 0\).

\((4 + m)^2 = 16\).

\(4 + m = \\pm 4\).

Solving gives \(m = 0\) or \(m = -8\).

Since \(m\) must be non-zero, \(m = -8\).

Substitute \(m = -8\) back into the quadratic: \(x^2 + 4x + 4 = 0\).

Factor: \((x + 2)^2 = 0\).

\(x = -2\).

Substitute \(x = -2\) into the curve equation to find \(y\):

\(y = (-2)^2 - 4(-2) + 4 = 4 + 8 + 4 = 16\).

Thus, the coordinates are \((-2, 16)\).

(i) To find the value of \(k\), we equate the line and the curve:

\(\frac{x^2}{4} = \frac{x}{k} + k\)

Rearranging gives:

\(kx^2 - 4x - 4k^2 = 0\)

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

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

Substituting \(a = k\), \(b = -4\), \(c = -4k^2\), we get:

\((-4)^2 - 4(k)(-4k^2) = 0\)

\(16 + 16k^3 = 0\)

\(k^3 = -1\)

\(k = -1\)

(ii) Substitute \(k = -1\) into the line equation:

\(y = -x - 1\)

Substitute into the curve equation \(4y = x^2\):

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

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

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

\(x = -2\)

Substitute \(x = -2\) back into \(y = -x - 1\):

\(y = -(-2) - 1 = 1\)

Thus, the coordinates of \(P\) are \((-2, 1)\).

To find the intersection points, substitute x = k - 2y into the curve equation xy = 6:

\((k - 2y)y = 6\)

\(2y^2 - ky + 6 = 0\)

This is a quadratic in y. For two distinct points, the discriminant must be positive:

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

Here, \(a = 2\), \(b = -k\), \(c = 6\).

\((-k)^2 - 4 \cdot 2 \cdot 6 > 0\)

\(k^2 - 48 > 0\)

\(k^2 > 48\)

\(k < -\sqrt{48} \text{ and } k > \sqrt{48}\)

To find the points of intersection, set the equations equal: \(3x - 2k = x^2 - kx + 2\).

Rearrange to form a quadratic equation: \(x^2 - kx - 3x + 2 + 2k = 0\).

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

Use the discriminant \(b^2 - 4ac\) to determine if there are real solutions: \(b = -(k+3), a = 1, c = 2 + 2k\).

Calculate the discriminant: \((k+3)^2 - 4(1)(2 + 2k)\).

Simplify: \((k+3)^2 - 8 - 8k = (k-1)^2\).

Since \((k-1)^2 \geq 0\) for all \(k\), the discriminant is non-negative, ensuring real solutions.

Hence, the line and the curve meet for all values of \(k\).

To find the value of k for which the line y = 6x + k is tangent to the curve y = 7/√x, we need to equate the derivatives and solve for k.

The derivative of y = 7/√x is:

\(\frac{d}{dx} \left( 7x^{-1/2} \right) = -\frac{7}{2}x^{-3/2}\)

The derivative of y = 6x + k is:

\(\frac{d}{dx} (6x + k) = 6\)

Setting the derivatives equal gives:

\(-\frac{7}{2}x^{-3/2} = 6\)

Solving for x:

\(x = \frac{49}{144}\)

Substitute x back into the curve equation to find y:

\(y = \frac{49}{12}\)

Substitute x and y into the line equation:

\(\frac{49}{12} = 6 \left( \frac{49}{144} \right) + k\)

Solve for k:

\(k = \frac{49}{24} \text{ or } 2.04\)