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Hey there! Thanks for pointing this out! We'll fix this as soon as we can.

Hey there! For organic chemistry reactions, we only generally write the major or desired products. In this case, this is a substitution reaction, so we only wrote the substitution product. If we include the other products, the $Na^{+}$ from $CH_3ONa$ will combine with $Br^{-}$ from $CH_3Br$ to form NaBr. I hope this clears things up!

Hey there! That’s fine. Doing it that way will lead to the same final molecule. It can be used as a "shortcut" to draw the product. Just keep in mind that there's an implicit hydrogen in the original molecule, which is eliminated during the formation of the carboxylic acid. Hope that clears things up!