[noda]

In the next scheme there are Four different modes of epoxide opening, one of them is the product,
acordding to the article exo cyclization would dominate over endo cyclization and that five-membered-ring formation would be faster than four-membered-ring formation
1.according to baldwin indeed exo cyclization of five-membered-ring would dominate over endo cyclization but I don't understand why?
2.Is there a possibility of formation of six-membered-ring?

[movies]

Baldwin's rules are pretty much empirically derived, although you can usually rationalize them if you build models.  Most of the disfavored cases require significant bond contortions in order to get the proper orbital alignment.  The disfavored cases are still possible, just a lot harder to do.

[noda]

does K2CO3 soluble in methanol? if it does can the anion carbonate take proton from methanol or from reagent (number 4) otherwise how does the ring closure occur?

[movies]

K₂CO₃ is sorta soluble in MeOH, generally MeOH is the solvent for these reactions, so you can usually get all the solid into solution.

I think in this case you probably generate a small amount of the deprotonated substrate which reacts very rapidly in an intramolecular fashion.  The MeOH may just be there for dissolving the carbonate base more effectively and maybe methoxide is the species that deprotonates the substrate.