How to find the strongest base using stability arguments
A vs. B

I will label your first reaction scheme, using the letters A-E for the deprotonated (i.e. as shown in the problem) form and the symbol $\ce{AH+}$ etc for the protonated form (after acting as a base and accepting a proton):

$$\ce{A (stronger base) + BH+ (stronger acid) &lt;=&gt; AH+ (weaker acid) + B (weaker base)}$$

Your argument about aromaticity is great; imidazolium ($\ce{BH+}$) is the stronger acid compared to cyclohexanol ($\ce{AH+}$), so the alkoxide ($\ce{A}$) is the stronger base compared to imidazole.

A vs. D

The protonated tetrahydrofuran (D) is highly reactive ("unstable") in water because it would want to get rid of a proton. That does not make it a stronger base. In your reaction scheme, you add another proton to a species that already is a strong acid (not sure if it even exists, see Why doesn&#39;t H₄O&#178;⁺ exist?).

$$\ce{A (stronger base) + DH+ (stronger acid) &lt;=&gt; AH+ (weaker acid) + D (weaker base)}$$

The cyclohexoxide ($\ce{A}$) is the stronger base than protonated tetrahydrofuran ($\ce{B}$) because we know phenol ($\ce{AH+}$) exists, but we don't think twice-protonated tetrahydrofuran ($\ce{BH+}$) exists.


  the O- one is more stable (on the left side), thus it is the weaker base, meaning the OH+ one (on the right side) is the stronger base


You have to specify stable in terms of what reaction. Protonated tetrahydrofuran is very stable against protonation, so it is not a base.


  However, when I compare the two acids on each side. I see that the OH one (right side) is more stable, thus it must be the weaker acid, so its conjugate base which is the O- one must be stronger.


Now you are comparing the two in terms of either one losing a proton, which is the right thing to consider when comparing two acids.

C and E

You are right about E, the nitrogen already makes four bonds, so it can't accept another proton. I'm not sure where a proton would go in C. In any case, it is not a good base.