Answer
$T_1=(\frac{m_1m_3}{m_1+m_2+m_3})g$; $T_2=[\frac{m_3(m_1+m_2)}{m_1+m_2+m_3}]g$
Work Step by Step
We can find the required tension in the strings as follows:
For block 1 $\Sigma F=0$
$\implies T_1-m_1a=0$.....eq(1)
For block 2 $\Sigma F=0$
$\implies -T_1+T_2-m_2a=0$....eq(2)
For block 3 $\Sigma F=0$
$\implies -T_2+m_3g-m_3a=0$......eq(3)
Adding eq(1),eq(2) and eq(3), we obtain:
$m_3g=(m_1+m_2+m_3)a$
This simplifies to:
$a=\frac{m_3g}{m_1+m_2+m_3}$
We plug in this value of $a$ in eq(1) to obtain:
$T_1=(\frac{m_1m_3}{m_1+m_2+m_3})g$
This is the tension between $m_1$ and $m_2$.
Now we find the tension between $m_2$ and $m_3$, that is:
$T_2=m_3(g-a)$
$\implies T_2=m_3(g-\frac{m_3g}{m_1+m_2+m_3})$
$T_2=[\frac{m_3(m_1+m_2)}{m_1+m_2+m_3}]g$
