Stability of cyclic radicals
Radicals are electron-deficient compounds. Therefore, according to Bent’s rule a radical should always populate an atomic orbital with a maximised $\mathrm p$ contribution. Ignoring 2 and 4 for a second, the radical occupies an $\mathrm{sp^2}$ orbital in 3 and 5 while it occupies an $\mathrm{sp^3}$ orbital in 1.[1] This already gives us $3\approx5&lt;1$.

In a next step, let’s consider 2 and 4. These, as you noticed, are adjacent to a &#x3c0; bond, so the radical can occupy a $\mathrm p$ type orbital and participate in &#x3c0; resonance. This is the main way to differentiate 1, 2 and 4, since all three could essentially adopt a planar configuration with the radical confined to a $\mathrm p$ orbital — yet in 2 it can also resonance with the allyl system and in 4 it can resonate across two double bonds. Thus, we now intermediately have $3\approx5&lt;1&lt;2&lt;4$.

All that remains is to determine the difference between 3 and 5, which is not trivial. In 3, you have a single double bond. This structure can distort a little bit, increasing the bond angle on the radical-bearing carbon towards something closer to $\mathrm{sp}$. The freely rotatable single bonds making up most of the ring allow this distortion. The phenyl radical can distort much less, since its geometry is basically already defined from the six-membered aromatic ring. Thus, we may assume 3 to be more stable than 5. Therefore: $5&lt;3&lt;1&lt;2&lt;4$.



Note:

[1]: In fact, rehybridisation is also possible in 1, giving the radical a pure $\mathrm p$ orbital and making the arrangement of substituents planar around the radical centre.