Here $\mathrm{T}_{0}=8 \mathrm{g}=80 \mathrm{N}$
$\mathrm{Also}, \mathrm{T}_{0}=2 \mathrm{T}$
$\therefore \mathrm{T}=40 \mathrm{N}$
From force diagram of $5 \mathrm{kg}$ block,
$50-T=5 a$
$50-40=5 a$
$a=2 m / s^{2}$
From force diagram of $m_{1}$
$\mathrm{T}-\mathrm{m}_{1} \mathrm{g}=\mathrm{m}_{1} \mathrm{a}$
or $40-m_{1} g=m_{1} \times 2$
or $40=10 \mathrm{m}_{1}+2 \mathrm{m}_{1}=12 \mathrm{m}_{1}$
$\therefore m_{1}=\frac{40}{12}=\frac{10}{3} \mathrm{kg}$.