From the line \(3x - y + 8 = 0\) we have \(y = 3x + 8\).

Substitute into the circle:

\[ x^2 + (3x + 8)^2 - 18x - 10(3x + 8) + 16 = 0. \]

Expand and simplify:

\[ x^2 + (9x^2 + 48x + 64) - 18x - 30x - 80 + 16 = 0 \;\Rightarrow\; 10x^2 + 0x + 0 = 0. \]

This gives the quadratic \(10x^2 = 0\), so

\[ x = 0 \quad(\text{a repeated root}). \]

Since the quadratic has a repeated root (discriminant \(b^2 - 4ac = 0\)), the line is tangent and the single point of intersection is obtained by \(y = 3x + 8\):

\[ y = 3(0) + 8 = 8. \]

Point of contact: \(\boxed{(0,\,8)}\).

For a tangent, the line and the curve intersect at exactly one point. Hence, substituting \( y = kx - 6 \) into \( y = x^2 + x - 2 \):

\[ kx - 6 = x^2 + x - 2 \]

Rearrange to form a quadratic in \(x\):

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

For tangency, the discriminant must be zero:

\[ b^2 - 4ac = 0 \]

\[ (1 - k)^2 - 4(1)(4) = 0 \]

Simplify:

\[ (1 - k)^2 = 16 \]

Take square roots:

\[ 1 - k = \pm 4 \]

Hence, the two possible values of \(k\) are:

\[ k = -3 \quad \text{or} \quad k = 5. \]

Final Answer: \(\boxed{k = -3 \text{ or } k = 5.}\)

From the line equation \(x + y = k\), we can write \(y = k - x.\)

Substitute \(y = k - x\) into the circle equation:

\[ x^2 + (k - x)^2 + 6x + 2(k - x) - 8 = 0 \]

Expand and simplify:

\[ x^2 + (k^2 - 2kx + x^2) + 6x + 2k - 2x - 8 = 0 \]

\[ 2x^2 + (-2k + 4)x + (k^2 + 2k - 8) = 0 \]

For tangency, the discriminant must be zero:

\[ b^2 - 4ac = 0 \]

\[ (-2k + 4)^2 - 4(2)(k^2 + 2k - 8) = 0 \]

Simplify:

\[ (4k^2 - 16k + 16) - 8(k^2 + 2k - 8) = 0 \]

\[ 4k^2 - 16k + 16 - 8k^2 - 16k + 64 = 0 \]

\[ -4k^2 - 32k + 80 = 0 \]

Divide by \(-4\):

\[ k^2 + 8k - 20 = 0 \]

Solve for \(k\):

\[ k = \frac{-8 \pm \sqrt{8^2 - 4(1)(-20)}}{2} = \frac{-8 \pm \sqrt{64 + 80}}{2} = \frac{-8 \pm \sqrt{144}}{2} \]

\[ k = \frac{-8 \pm 12}{2} \]

Hence,

\[ k = 2 \quad \text{or} \quad k = -10. \]

Final Answer: \(\boxed{k = 2 \text{ or } k = -10.}\)