A boy stands on a platform that can freely rotate about its...

Let $I_1$ be the moment of inertia of the platform plus boy, and $I_2$ be the moment of inertia of the umbrella.

$I_1 = 3.0 \times 10^{-3} \text{ kg-m}^2$

$I_2 = 2.0 \times 10^{-3} \text{ kg-m}^2$

Let $\omega_1$ be the angular velocity of the platform relative to the ground. The angular velocity of the umbrella relative to the platform is $\omega_{rel} = 2.0 \text{ rev/s}$.

The angular velocity of the umbrella relative to the ground is $\omega_2 = \omega_1 + \omega_{rel}$.

Since no external torque acts on the system about the vertical axis, the total angular momentum is conserved.

$L_i = L_f$

$0 = I_1 \omega_1 + I_2 \omega_2$

$0 = I_1 \omega_1 + I_2 (\omega_1 + \omega_{rel})$

$(I_1 + I_2) \omega_1 = -I_2 \omega_{rel}$

$\omega_1 = -\dfrac{I_2 \omega_{rel}}{I_1 + I_2}$

Substituting the given values:

$\omega_1 = -\dfrac{2.0 \times 10^{-3} \times 2.0}{3.0 \times 10^{-3} + 2.0 \times 10^{-3}}$

$\omega_1 = -\dfrac{4.0 \times 10^{-3}}{5.0 \times 10^{-3}}$

$\omega_1 = -0.8 \text{ rev/s}$

The negative sign indicates that the platform rotates in the direction opposite to the umbrella's rotation. The magnitude of the angular velocity imparted to the platform is $0.8 \text{ rev/s}$.