Answer
$a_1 = \frac{2m_2~g}{4m_1 + m_2}$
$a_2 = \frac{m_2~g}{4m_1 + m_2}$
Work Step by Step
Let $a_1$ be the acceleration of block 1.
Let $a_2$ be the acceleration of block 2.
When block 2 falls a distance of one unit, we can see that block 1 will move to the right 2 units. Therefore, $a_1 = 2a_2$.
Let $T_1$ be the tension in the string on block 1.
Let $T_2$ be the tension in the string on block 2.
When we examine the pulley above block 2, we can see that $T_2 = 2T_1$.
$T_1 = m_1 ~a_1$
$m_2~g - T_2 = m_2~a_2$
$m_2~g - 2T_1 = m_2~a_2$
$m_2~g - 2m_1~a_1 = m_2~a_2$
$m_2~g - 4m_1~a_2 = m_2~a_2$
$m_2~g = 4m_1~a_2 + m_2~a_2$
$a_2 = \frac{m_2~g}{4m_1 + m_2}$
$a_1 = 2a_2 = \frac{2m_2~g}{4m_1 + m_2}$
