The portion of the line $$4x + 5y = 20$$ in the first quadrant is trisected by the lines $$L_1$$ and $$L_2$$ passing through the origin. The tangent of an angle between the lines $$L_1$$ and $$L_2$$ is :

The line $$4x + 5y = 20$$ intersects the axes at $$A(5, 0)$$ and $$B(0, 4)$$.

The portion AB is trisected, giving two points that divide AB into three equal parts.

The trisection points are:

$$P_1 = A + \frac{1}{3}(B - A) = \left(5 - \frac{5}{3}, 0 + \frac{4}{3}\right) = \left(\frac{10}{3}, \frac{4}{3}\right)$$

$$P_2 = A + \frac{2}{3}(B - A) = \left(5 - \frac{10}{3}, 0 + \frac{8}{3}\right) = \left(\frac{5}{3}, \frac{8}{3}\right)$$

Lines through origin:

$$L_1$$: passes through origin and $$P_1\left(\frac{10}{3}, \frac{4}{3}\right)$$. Slope $$m_1 = \frac{4/3}{10/3} = \frac{2}{5}$$.

$$L_2$$: passes through origin and $$P_2\left(\frac{5}{3}, \frac{8}{3}\right)$$. Slope $$m_2 = \frac{8/3}{5/3} = \frac{8}{5}$$.

The tangent of the angle between $$L_1$$ and $$L_2$$:

$$\tan\theta = \left|\frac{m_2 - m_1}{1 + m_1 m_2}\right| = \left|\frac{\frac{8}{5} - \frac{2}{5}}{1 + \frac{2}{5} \cdot \frac{8}{5}}\right| = \left|\frac{\frac{6}{5}}{1 + \frac{16}{25}}\right| = \left|\frac{\frac{6}{5}}{\frac{41}{25}}\right| = \frac{6}{5} \times \frac{25}{41} = \frac{30}{41}$$

The answer is $$\frac{30}{41}$$, which corresponds to Option (4).