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The key to solve this question is to calculate the moment of inertia of this trimmer. There are two components, one is the wire spool and the other is the plastic wire sticking out from the spool. We can use the following diagram to show how the trimmer can be considered in terms of components of simple structures/shapes.

The next step is to find the masses of these different objects. The mass of the wires can be obtained by multiplying the length and the linear density (10g/m).

The mass of the trimmer spool is a bit challenging. First we treat this as uniform disks so we can get its area density of mass (defined as ||\lambda||) using the 0.1kg divided by the area of the circular ring: $$\lambda=\frac{0.1}{\pi R^2-\pi r^2}.$$

To find M, you multiple the area density with the area of the large disk ||M=\lambda \cdot \pi R^2||.

Also you will have ||m=\lambda \cdot \pi R^2||.

Then you can calculate the total moment of inertia using ideas shown in the diagram.

Once you have ||I_{total}||, you can find its final rotational kinetic energy ||K_{rot}=\frac {1}{2}I_{total}\omega ^2||.

The (b) part of the question is simpler. First you get the torque ||\tau=fL_1||. Since ||dW=\tau d\theta||, you can find: $$P=\frac{dW}{dt}=\tau \frac {d\theta}{dt}=\tau \omega .$$