In seals the gene for the length of the whiskers has two alleles. The dominant a

The offspring will have long whiskers

WW Ww WW Ww

As you can see, all of them are long-whiskered.

I personally like to write down the info given in the question on my paper first.  So I start by writing:
W = allele for long whiskers
w = allele for short whiskers
A homozygous dominant seal would be "WW" (homozygous dominant = 2 CAPITAL letters).
A heterozygous seal would be "Ww" (heterozygous = 1 CAPITAL & 1 lowercase).
The cross is in the question therefore: WW x Ww.

The Punnett Square would look like this:

http://leavingbio.net/Genetics-higher%20level_files/image034.jpg

The possible gametes from the homozygous parent seal are on the left in front of the rows, & the possible gametes from the heterozygous parent are above the columns. We fill in the boxes by copying "one letter from the left, one letter from the top".

Analyzing our results, we find that 50% of our offspring (2 of 4 boxes) are "WW", and 50% (2 of 4 boxes) are "Ww". In terms of phenotype (what they would look like) 100% would have long whiskers (because all of the offspring have at least one "W", which codes for long whiskers).

So the answer to question 1a is: 0% would have short whiskers.  The only way to have short whiskers is to be "ww", and that combo is not possible from the parents in this cross.