In the diagram below, the graphs of f(x) = a/x + q and g(x) = -x + 4 are drawn. Graphs of f and g intersect at A and C. f and g intersect the x-axis at point E and D respectively. B lies on the y-axis. The horizontal asymptote of f intersect g at point B.

Point B lies on the y-axis, so its x-coordinate is 0.
Point B is also the intersection of the horizontal asymptote of $f$ and the graph of $g$.
The equation of $g(x)$ is given as $g(x) = -x + 4$.
Substitute $x=0$ into $g(x)$ to find the y-coordinate of B:

$$g(0) = -(0) + 4 = 4$$

So, the coordinates of B are $(0, 4)$.
Since B is on the horizontal asymptote of $f(x) = \frac{a}{x} + q$, and the horizontal asymptote is $y=q$, we can conclude that $q=4$.

The coordinates of B are $\boxed{(0, 4)}$.

The general equation for $f$ is $f(x) = \frac{a}{x} + q$.
From Step 1, we found $q=4$. So, $f(x) = \frac{a}{x} + 4$.
We are given that point P$(\frac{1}{4}; -4)$ lies on the graph of $f$. Substitute these coordinates into the equation for $f$:

$$-4 = \frac{a}{\frac{1}{4}} + 4$$ $$-4 = 4a + 4$$

Subtract 4 from both sides:

$$-4 - 4 = 4a$$ $$-8 = 4a$$

Divide by 4:

$$a = \frac{-8}{4}$$ $$a = -2$$

Therefore, the equation of $f$ is $f(x) = -\frac{2}{x} + 4$.

The equation of $f$ is $\boxed{f(x) = -\frac{2}{x} + 4}$.

Points A and C are the intersection points of $f(x)$ and $g(x)$.
Set $f(x) = g(x)$:

$$-\frac{2}{x} + 4 = -x + 4$$

Subtract 4 from both sides:

$$-\frac{2}{x} = -x$$

Multiply both sides by $-x$ (assuming $x \neq 0$):

$$2 = x^2$$

Take the square root of both sides:

$$x = \pm \sqrt{2}$$

From the graph, point A is in the second quadrant (negative x-value) and point C is in the first quadrant (positive x-value).
So, the x-coordinate of A is $-\sqrt{2}$ and the x-coordinate of C is $\sqrt{2}$.

The x-coordinates of A and C are $\boxed{x_A = -\sqrt{2} \text{ and } x_C = \sqrt{2}}$.

To calculate the area of $\triangle BOD$, we need the coordinates of B, O, and D.
O is the origin, so its coordinates are $(0, 0)$.
From Step 1, the coordinates of B are $(0, 4)$.
D is the x-intercept of $g(x)$. To find D, set $g(x) = 0$:

$$-x + 4 = 0$$ $$x = 4$$

So, the coordinates of D are $(4, 0)$.
$\triangle BOD$ is a right-angled triangle with the right angle at O.
The length of the base OD is the distance from $(0,0)$ to $(4,0)$, which is 4 units.
The length of the height OB is the distance from $(0,0)$ to $(0,4)$, which is 4 units.
The area of a triangle is given by the formula: Area $= \frac{1}{2} \times \text{base} \times \text{height}$.

$$\text{Area of } \triangle BOD = \frac{1}{2} \times OD \times OB$$ $$\text{Area of } \triangle BOD = \frac{1}{2} \times 4 \times 4$$ $$\text{Area of } \triangle BOD = \frac{1}{2} \times 16$$ $$\text{Area of } \triangle BOD = 8 \text{ square units}$$

The area of $\triangle BOD$ is $\boxed{8 \text{ square units}}$.