Hello bhkuk,
concerning your questions:
for 1) The sodiumhydride will deprotonate the Hydroxyl-group in your molecule. This is necessary, because it will create a much better nucleophile.
for 2) This is a common protocol. You generate benzyl-iodide in situ, wihich is far more reactive than the corresponding bromide.
Regarding the reaction time: Check a TLC after some hours. Perhaps the one reporting this protocol just didn't want to stay in the lab for too long, so he decided to go home and let stirr overnight? ;-)
for 3) You should avoid acidic conditions during the work up in any case. The acetonide protecting group would not be stable under these conditions. So, just a use a base for work-up.
kind regards
miraculix