1. Find the intercepts of the given line in the first quadrant

The line is $4x+5y=20.$

Its intercepts are:

• On the $x$-axis: set $y=0$, $4x=20 \Rightarrow x=5,$ so point $A=(5,0)$.
• On the $y$-axis: set $x=0$, $5y=20 \Rightarrow y=4,$ so point $B=(0,4)$.

Thus, the portion in the first quadrant is the segment joining $A(5,0)$ and $B(0,4)$.

2. Find the trisection points of segment $AB$

The lines $L_1$ and $L_2$ pass through the origin and trisect the segment $AB$, so they pass through the two internal trisection points of $AB$.

Using section formula from $A(5,0)$ to $B(0,4)$:

First trisection point $P$

This divides $AB$ in the ratio $1:2$. $P=\left(\frac{2\cdot 5+1\cdot 0}{3},\frac{2\cdot 0+1\cdot 4}{3}\right)=\left(\frac{10}{3},\frac{4}{3}\right).$

Second trisection point $Q$

This divides $AB$ in the ratio $2:1$. $Q=\left(\frac{1\cdot 5+2\cdot 0}{3},\frac{1\cdot 0+2\cdot 4}{3}\right)=\left(\frac{5}{3},\frac{8}{3}\right).$

3. Find slopes of the lines through the origin

Since $L_1$ and $L_2$ pass through the origin and points $P$ and $Q$ respectively:

$m_1=\frac{\frac{4}{3}}{\frac{10}{3}}=\frac{2}{5},$ $m_2=\frac{\frac{8}{3}}{\frac{5}{3}}=\frac{8}{5}.$

4. Use the formula for angle between two lines

If two lines have slopes $m_1$ and $m_2$, then $\tan\theta=\left|\frac{m_2-m_1}{1+m_1m_2}\right|.$

Substitute $m_1=\frac{2}{5}$ and $m_2=\frac{8}{5}$:

$\tan\theta=\left|\frac{\frac{8}{5}-\frac{2}{5}}{1+\left(\frac{2}{5}\cdot\frac{8}{5}\right)}\right|$

$=\frac{\frac{6}{5}}{1+\frac{16}{25}}$

$=\frac{\frac{6}{5}}{\frac{41}{25}}$

$=\frac{6}{5}\cdot\frac{25}{41}$

$=\frac{30}{41}.$

5. Check options

The correct value is $\boxed{\frac{30}{41}}.$

So the correct option is A.

6. Comparison with stored correct answer

Stored correct answer: A

Derived answer: A

They agree.