Given equations:

1. Circle: x² + y² + 6x - 8y = 0
2. Line: y = mx - 1/3
To find the points of intersection, we substitute the equation of the line (2) into the equation of the circle (1):

x² + (mx - 1/3)² + 6x - 8(mx - 1/3) = 0

Simplifying the equation, we get:
x² + m²x² - 2mx/3 + 1/9 + 6x - 8mx/3 + 8/3 = 0

Combining like terms:
(1 + m²)x² + (6 - 2m/3 - 8m/3)x + (1/9 + 8/3) = 0

Now, let's analyze the cases:

(i) For the line to intersect the circle at two distinct points, the quadratic equation formed by the above equation should have two distinct real roots. This happens when the discriminant is greater than zero:

Discriminant = (6 - 2m/3 - 8m/3)² - 4(1 + m²)(1/9 + 8/3) > 0

Simplifying the inequality:
36 - 4m + 8m + 64m²/9 - 32m/3 - 32m²/3 > 0
(64m² - 32m + 36)/9 > 0
64m² - 32m + 36 > 0

Solving this quadratic inequality, we find the range of values of m for which the line intersects the circle at two distinct points.

(ii) For the line to be tangent to the circle, the quadratic equation should have a repeated root. This occurs when the discriminant is equal to zero:

Discriminant = (6 - 2m/3 - 8m/3)² - 4(1 + m²)(1/9 + 8/3) = 0

Solving this equation will give us the value(s) of m for which the line is tangent to the circle.

(iii) If the discriminant is less than zero, there are no real roots, meaning the line does not intersect the circle.