To find the equation of the tangent to the circle that is parallel to the line 3x + 4y - 1 = 0, we need to follow these steps:

1. Find the center and radius of the given circle.
2. Find the slope of the given line.
3. Use the slope of the line and the center of the circle to find the slope of the tangent line.
4. Use the slope of the tangent line and the center of the circle to find the equation of the tangent line.

Let's go through these steps in detail:

1. Find the center and radius of the given circle:
The equation of the circle is x^2 + y^2 - 2x + 4y - 4 = 0. We can rewrite this equation in the standard form by completing the square for both x and y terms:

(x^2 - 2x) + (y^2 + 4y) = 4
(x^2 - 2x + 1) + (y^2 + 4y + 4) = 4 + 1 + 4
(x - 1)^2 + (y + 2)^2 = 9

From this equation, we can see that the center of the circle is (1, -2) and the radius is √9 = 3.

2. Find the slope of the given line:
The line 3x + 4y - 1 = 0 can be rewritten in slope-intercept form as y = (-3/4)x + 1/4. The slope of this line is -3/4.

3. Use the slope of the line and the center of the circle to find the slope of the tangent line:
Since the tangent line is parallel to the given line, it will have the same slope. Therefore, the slope of the tangent line is also -3/4.

4. Use the slope of the tangent line and the center of the circle to find the equation of the tangent line:
The equation of a line in point-slope form is given by y - y1 = m(x - x1), where (x1, y1) is a point on the line and m is the slope.

Using the slope -3/4 and the center of the circle (1, -2), we can write the equation of the tangent line as:
y - (-2) = (-3/4)(x - 1)
y + 2 = (-3/4)x + 3/4
y = (-3/4)x + 3/4 - 2
y = (-3/4)x - 5/4

So, the equation of the tangent line to the circle that is parallel to the line 3x + 4y - 1 = 0 is 3x + 4y = 5/4.

However, none of the given options match this equation exactly. It seems there might be a mistake in the options provided.