#### Answer

The acceleration of $m_1$ is $\frac{2m_2~g}{4m_1 + m_2}$

#### Work Step by Step

We can set up a force equation for $m_2$. Let $T$ be the tension in the rope. Therefore;
$\sum F = m_2~a_2$
$m_2~g - 2T = m_2~a_2$
$T = \frac{m_2~g - m_2~a_2}{2}$
We can use this expression for the tension $T$ in the force equation for $m_1$. Note that $a_2 = \frac{a_1}{2}$.
$\sum F = m_1~a_1$
$T = m_1~a_1$
$\frac{m_2~g - m_2~a_2}{2} = m_1~a_1$
$m_2~g - m_2~(\frac{a_1}{2}) = 2m_1~a_1$
$2m_2~g - m_2~a_1 = 4m_1~a_1$
$2m_2~g = 4m_1~a_1 + m_2~a_1$
$a_1 = \frac{2m_2~g}{4m_1 + m_2}$
The acceleration of $m_1$ is $\frac{2m_2~g}{4m_1 + m_2}$.