Answer
$L_1 = 2.34~m$
$L_2 = 1.66~m$
Work Step by Step
We can write an expression for the speed along string 1 as:
$v_1 = \sqrt{\frac{F_T}{\mu_1}}$
We can find the time $t_1$ for the pulse to reach the end of string 1.
$t_1 = \frac{L_1}{v_1}$
$t_1 = \frac{L_1}{\sqrt{\frac{F_T}{\mu_1}}}$
$t_1 = L_1~\sqrt{\frac{\mu_1}{F_T}}$
Note that $\mu_2 = 2\mu_1$. We can write an expression for the speed along string 2.
$v_2 = \sqrt{\frac{F_T}{\mu_2}}$
$v_2 = \sqrt{\frac{F_T}{2\mu_1}}$
We can find the time $t_2$ for the pulse to reach the end of string 2.
$t_2 = \frac{L_2}{v_2}$
$t_2 = \frac{L_2}{\sqrt{\frac{F_T}{2\mu_1}}}$
$t_2 = \sqrt{2}~L_2~\sqrt{\frac{\mu_1}{F_T}}$
If the pulses reach the end at the same time, then $t_1=t_2$.
$L_1~\sqrt{\frac{\mu_1}{F_T}} = \sqrt{2}~L_2~\sqrt{\frac{\mu_1}{F_T}}$
$L_1 = \sqrt{2}~L_2$
We also know that the sum of the two lengths is $4.0~m$:
$L_1+L_2 = 4.0~m$
$\sqrt{2}~L_2+L_2 = 4.0~m$
$L_2 = \frac{4.0~m}{\sqrt{2}+1}$
$L_2 = 1.66~m$
We can find $L_1$:
$L_1+L_2 = 4.0~m$
$L_1 = 4.0~m - L_2$
$L_1 = 4.0~m - 1.66~m$
$L_1 = 2.34~m$
