Answer
The mass $m_2$ hangs below the ceiling a distance of $L_1+\frac{(m_1+m_2)~g}{k_1}+L_2+\frac{(m_2)~g}{k_2}$
Work Step by Step
The force of the second spring is equal to the weight of the hanging mass $m_2$. Since the second spring pulls down on the hanging mass $m_1$ with the weight of the hanging mass $m_2$, the force of the first spring is equal to the sum of the weights of both hanging masses.
We can find the distance $x_1$ that the first spring stretches.
$k_1~x_1 = (m_1+m_2)~g$
$x_1 = \frac{(m_1+m_2)~g}{k_1}$
We can find the distance $x_2$ that the second spring stretches.
$k_2~x_2 = (m_2)~g$
$x_2 = \frac{(m_2)~g}{k_2}$
We can find the total distance $d$ that the mass $m_2$ hangs below the ceiling.
$d = L_1+x_1+L_2+x_2$
$d = L_1+\frac{(m_1+m_2)~g}{k_1}+L_2+\frac{(m_2)~g}{k_2}$
The mass $m_2$ hangs below the ceiling a distance of $L_1+\frac{(m_1+m_2)~g}{k_1}+L_2+\frac{(m_2)~g}{k_2}$
