Let's see how we can start breaking this problem down into easier parts.

First, I'm going to label some points just to make my explanation easier. Label the point on each disc furthest from the other disc $A$ and $B$, and label the points midway on the straight sections of the band $C$ and $D$ (it doesn't really matter what way round this is done for either by the symmetry of the problem). See my (badly drawn) diagram here.

Now, by the symmetry of the problem, we can consider the "top" part of the band $ACB$ and the "bottom" part of the band $ADB$ separately. Each part has natural length $l=\pi r$ and extension $x=6r$. Using the normal Hooke's law formula with modulus $\lambda = \pi mg/12$, the tension in the top part is given by $$T=\frac{\lambda x}{l}=\frac{(\pi mg/12)(6r)}{\pi r}=\frac{1}{2}mg.$$ The tension in the bottom part is also given by $T$ by symmetry, so the total tension acting on one of the discs is simply $2T=mg$. Hopefully that should give you an idea how to do part (i).

The initial total elastic potential energy in the band is easier to calculate and we don't need to split the band anymore. Using the normal formula with $x=12r$ and $l=2\pi r$, we get $$E_{\text{P}}=\frac{\lambda x^2}{2l}=\frac{(\pi mg/12) (12r)^2}{2(2 \pi r)}=3mgr.$$ Perhaps, you can now calculate the elastic potential energy at the point just before the two discs collide and subsequently do part (ii).

For a particle held between two springs, you can just work out the tension in each spring separately and then, if the particle is in equilibrium, the resultant tension must vanish. I'm not really sure what else I can say without looking at a specific problem.

I hope this is helpful and I haven't confused you too much.