Is the amount of #Na(OH)# with respect to amide in hydrolysis important?

Yes. With acid-catalyzed hydrolysis towards the carboxylic acid, it is generally easier, but with base-catalyzed hydrolysis towards the carboxylate, it requires that #"OH"^(-)# is a good enough nucleophile, rather than relying on making the carbonyl a better electrophile by protonating it.

Amides are not the most reactive carbonyl species (thank god, or else we'd turn to mush; how about them peptide bonds?), so using #"OH"^(-)# as a nucleophile can be difficult in this case.

In accordance with Le Chatelier's principle, the more #"OH"^(-)# you have, the faster the first step of the reaction is and thus the more easily the reaction can proceed.

The mechanism is:

The first step, as we mentioned, is difficult, so we must add excess #"NaOH"# to drive the reaction forward.

The second step is also difficult; forming #"NH"_2^(-)# is hard because its conjugate acid is #"NH"_3#, whose #"pKa"# is #36#, and thus is quite basic. So, you can imagine that #"NH"_2^(-)# is even more basic, and it would normally rather stay on the amide.

Once we get past that, #"NH"_2^(-)# easily grabs a proton from the carboxylic acid and generates the quite unreactive carboxylate product.