Writing equation from the reference frame of pulley
\(\frac{3 m_1 g}{2}-T=m_1 a \ldots(i)\)
\(T-\frac{3 m_2 g}{2}=m_2 a \ldots(ii)\)
Add \((i)\) and \((ii)\)
\(\frac{3 g}{2}\left(\frac{m_1-m_2}{m_1+m_2}\right)=a\)
Use \(a\) in eq. \((i)\) or \((ii)\)
Solving \(T=\frac{3 m_1 m_2}{m_1+m_2} g\)
$\left[g=10 m / s ^{2}\right]$.