Note : In all acid-base reactions, the end products are always the weaker base and weaker acid. This is due to equilibrium considerations.
1. Ethyl alcohol would first get protonated by HCl as it is a strong acid and -OH is an acid-labile group. Thus the protonated intermediate and Cl- would form, which would lead to a nucleophilic substitution giving ethyl chloride as end product, but that's a different story.
2. I see no reason why the methyl group would leave like that...not sure about this one. Possibly so.
3. Imidazole is amphoteric..the protonation would occur at the nitrogen which is at 3-position, or the nitrogen having a double bond. Doesn't look like it will have more than 3 resonance forms.
4. Dipole moment is a vector, being the product of charge(scalar) and distance(vector). The size of fluorine is very small, decreasing the distance factor considerably. But the size of chlorine is large compared to fluorine, so even though the negative charge on fluorine would be strong, methyl chloride has the larger dipole moment. This is the easiest way I could think of to explain it to you. I don't remember the actual reasons.
5. Check the stability of the conjugate base that would form, by removing protons from the required locations. If we remove a proton from O, resonance occurs freely between the two O atoms. This delocalises the negative charge, in turn stabilizing the conjugate base. However if we remove a proton from the methyl group, there is no scope for resonance and the base is not stable in any way (though COOH provides a -I effect in the second case, resonance in the first case is a stronger stabilizer). Thus the proton attached to O is more acidic.
6. Water is amphoteric...it would readily react in an acid base reaction as tert-butoxide anion is not stabilized by any factor.
Disclaimer : I do not guarantee that my answers are in any way correct.