The magnitude of acceleration of both the blocks will be same and is given by

$a=\frac{4 g \sin {60}^{\circ }-g \sin {30}^{\circ }}{4+1}$

$=\frac{20\sqrt{3}-5}{5}$

$a=\left(4\sqrt{3}-1\right)$ $\mathrm{m} {\mathrm{s}}^{-1}$

Applying Newton's second law for second block (along the direction parallel to the incline on which it is placed)

$T-5=1\times a$

$\Rightarrow T=5+4\sqrt{3}-1$

$=4\left(\sqrt{3}+1\right) \mathrm{N}$