A small block of mass m = 0.50 kg lies on a rough horizontal turntable…

Step 1. The condition for rest. The spring pulls the block inwards with $k(r - l_0)$ (it pushes outwards if $r < l_0$). Circular motion needs an inward force $m\omega^2 r$. Friction must make up the difference, and it can be at most $\mu mg$ either way:

Since $k > m\omega^2$ here, this gives the band

At the inner end the block is about to slide outwards, and friction pulls it in. At the outer end it is about to slide inwards, and friction pushes it out.

Step 2. Numbers for (a). $kl_0 = 15\ \text{N}$, $\mu mg = 2.0\ \text{N}$ and $k - m\omega^2 = 50\ \text{N/m} - 18\ \text{N/m} = 32\ \text{N/m}$. So

Step 3. Speeding up (b). At $r = l_0$ the spring does nothing, so friction alone supplies $m\omega^2 l_0$. The block starts to slide when $m\omega^2 l_0 = \mu mg$:

As the table speeds up further, the block creeps outwards, always on the point of sliding out. So it stays at the inner end of the band. At $6.0\ \text{rad/s}$ that is $r = 0.406\ \text{m}$.

Step 4. Slowing down (c). As $\omega$ falls, the whole band moves inwards. The block, now at $0.406\ \text{m}$, tends to slide inwards. It stays where it is until the outer end of the band comes down to it, when $\dfrac{17\ \text{N}}{k - m\omega^2} = 0.406\ \text{m}$, at $\omega = 4.04\ \text{rad/s}$. Below that it creeps inwards, staying at the outer end of the band. At $3.0\ \text{rad/s}$: