we draw the free body diagram of the system (force diagram) and we note that the acceleration is towards the heavier mass as shown below; (since \(8 kg >5 kg\) )
for the \(8 kg\) mass
\(\Sigma F=m a\) (Newton's second law)
\(m g \sin \theta-T=m_{2} a\)
\(8 g \sin 35-T=8 a \ldots\) (1)
for the \(5 kg\) mass
\(T-m_{1} g \sin \theta=m_{1} a\)
\(T-5 g \sin 35=5 a \ldots(2)\)
combining (1) and (2), we obtain
\(8 g \sin 35-5 g \sin 35=8 a+5 a\)
\(3 g \sin 35^{\circ}=13 a\)
\(a=\frac{3 g \sin 35}{13}=0.13 g\)