In the arrangement shown in the figure, the tension in the string...

Taking $g = 10 \text{ m/s}^2$.

For the $2 \text{ kg}$ block, the equation of motion is:
$T - \mu_1 m_1 g = m_1 a$

Substituting the given values ($T = 16 \text{ N}$, $m_1 = 2 \text{ kg}$, $a = 0.5 \text{ m/s}^2$):
$16 - \mu_1 (2)(10) = 2(0.5)$
$16 - 20\mu_1 = 1$
$20\mu_1 = 15 \Rightarrow \mu_1 = 0.75$

For the $4 \text{ kg}$ block, the component of weight along the incline is m_2 g \sin 30^{}^{{}^{\circ}} = 4 \times 10 \times 0.5 = 20 \text{ N}.
Since $20 \text{ N} > 16 \text{ N}$, the block moves down the incline.

The equation of motion for the $4 \text{ kg}$ block is:
m_2 g \sin 30^{}^{{}^{\circ}} - T - \mu_2 m_2 g \cos 30^{}^{{}^{\circ}} = m_2 a

Substituting the given values ($m_2 = 4 \text{ kg}$, $a = 0.5 \text{ m/s}^2$):
4(10)\sin 30^{}^{{}^{\circ}} - 16 - \mu_2(4)(10)\cos 30^{}^{{}^{\circ}} = 4(0.5)
$20 - 16 - 40\mu_2 \left(\dfrac{\sqrt{3}}{2}\right) = 2$
$4 - 20\sqrt{3}\mu_2 = 2$
$20\sqrt{3}\mu_2 = 2$
$\mu_2 = \dfrac{1}{10\sqrt{3}} \approx 0.0577 \approx 0.06$

Therefore, $\mu_1 = 0.75$ and $\mu_2 = 0.06$.

Answer: $\mu_1 = 0.75$, $\mu_2 = 0.06$