The graphs of f(x) = ax2 + bx + c and g(x) = -4/(x-3) + 8 are drawn below. P is the turning point of f and the point of intersection of the asymptotes of g. M and T are x-intercepts of f and g respectively.

The function $g(x) = \frac{-4}{x-3} + 8$ is a rational function. The denominator cannot be zero.

The domain of $g$ is all real numbers except $x=3$.
The domain of $g$ is $\boxed{x \in \mathbb{R}, x \neq 3}$.

The function $f(x)$ is a parabola with turning point P $(3; 8)$. From the graph, the parabola opens downwards, meaning the turning point is a maximum.
Therefore, the maximum value of $f(x)$ is the y-coordinate of the turning point.
The range of $f$ is $\boxed{y \leq 8}$.

From the graph, the function $g(x)$ is below or equal to $f(x)$ between the vertical asymptote $x=3$ and the intersection point D $(5; 6)$.
Since $x=3$ is an asymptote for $g(x)$, $x$ cannot be equal to 3. At point D, $x=5$, $g(5)=f(5)=6$, so $x=5$ is included.
The values of $x$ for which $g(x) \leq f(x)$ are $\boxed{3 < x \leq 5}$.

First, we need the equation of $f(x)$. The turning point is P $(3; 8)$, so $f(x) = a(x-3)^2 + 8$.
The graph passes through D $(5; 6)$. Substitute D into the equation:

So, $f(x) = -\frac{1}{2}(x-3)^2 + 8$.
Now, we solve $f(x) < 6$:

Multiply by $-2$ and reverse the inequality sign:

Take the square root of both sides:

This implies two separate inequalities:

The values of $x$ for which $f(x) < 6$ are $\boxed{x < 1 \text{ or } x > 5}$.

From Step 4, we found $f(x) = -\frac{1}{2}(x-3)^2 + 8$.
Expand the expression:

The equation is shown to be $f(x) = \boxed{-\frac{1}{2}x^2 + 3x + \frac{7}{2}}$.

M is the x-intercept of $f(x)$, and T is the x-intercept of $g(x)$.
To find T, set $g(x) = 0$:

So, T is $(\frac{7}{2}; 0)$.

To find M, set $f(x) = 0$:

Multiply by 2 to clear fractions:

Factor the quadratic equation:

From the graph, M is the negative x-intercept, so M is $(-1; 0)$.
The length of MT is the distance between the x-coordinates of M and T:

The length of MT is $\boxed{\frac{9}{2} \text{ units}}$.

The point D is $(5; 6)$.
To find the equation of the tangent line, we need its slope, which is the derivative of $f(x)$ evaluated at $x=5$.

Find the derivative $f'(x)$:

Now, evaluate $f'(x)$ at $x=5$ to find the slope $m$:

Using the point-slope form of a linear equation, $y - y_1 = m(x - x_1)$, with D $(5; 6)$ and $m = -2$:

The equation of the tangent to $f$ at D is $\boxed{y = -2x + 16}$.