Step 1: Identify the centre and radius of the circle.
The circle $(x-1)^2 + (y+2)^2 = 4$ has centre $C = (1, -2)$ and radius $r = 2$.
Step 2: State the key geometric fact.
When a tangent touches a circle, the radius to the point of tangency is perpendicular to the tangent. So the line from $C = (1, -2)$ to the point of contact $(\alpha, \beta)$ is perpendicular to $3x - 4y = 1$. Therefore $(\alpha, \beta)$ is the foot of the perpendicular from $C$ to the tangent line.
Step 3: Write the tangent line in standard form.
$3x - 4y - 1 = 0$. Here $a = 3$, $b = -4$, $c = -1$, and the reference point is $(x_1, y_1) = (1, -2)$.
Step 4: Compute $ax_1 + by_1 + c$.
$3(1) + (-4)(-2) + (-1) = 3 + 8 - 1 = 10$.
Step 5: Compute $a^2 + b^2$.
$3^2 + (-4)^2 = 9 + 16 = 25$.
Step 6: Apply the foot-of-perpendicular formula.
The foot $(\alpha, \beta)$ is \[ \alpha = x_1 - \frac{a(ax_1 + by_1 + c)}{a^2 + b^2} = 1 - \frac{3 \times 10}{25} = 1 - \frac{30}{25} = 1 - \frac{6}{5} = -\frac{1}{5} \] \[ \beta = y_1 - \frac{b(ax_1 + by_1 + c)}{a^2 + b^2} = -2 - \frac{(-4) \times 10}{25} = -2 + \frac{40}{25} = -2 + \frac{8}{5} = \frac{-10 + 8}{5} = -\frac{2}{5} \]
Step 7: State the final answer.
\[ \boxed{\alpha = -\dfrac{1}{5},\ \beta = -\dfrac{2}{5}} \]