2 parent dogs are each heterozygous for 2 traits: fur color & white spotting..?

I'm not 100% sure but i would think its B

The genes are sex linked. The parrents may bea llblack, but they could hold teh recessive genes for different colors.

or the mom could be getting ourd =P hahaha jk

This can only occur whereby there is an improper change in the process of reproduction, this is irregular but it can occur though not frequently so i guess the answer to your question is; genes for these traits mutate.

The answer is C.
As there are more than one pair for the fur color and spotting and they should be assort independently.

The question tells you that the parents are heterozygous for two traits, so this rules out answer a (the male parent can't be heterozygous, but could only be hemizygous, since there is only one X chromosome).  Generally in mammals, genes don't mutate frequently, so you can ignore answer b.  If the genes were on the same chromosome (and close together), then you would have a lot of puppies that look like the parents and a few recombinants that have different phenotypes from the parents.  Since the question didn't talk about this at all, then answer d is not the answer.  This leaves answer c as the correct answer.  Here is why:

If black is B and red is b, non-spotting is S and spotting is s, then the parents are:

BbSs  x  BbSs

The offspring would be in the following 9:3:3:1 ratio:

9 B-S-, black, non-spotted
3 bbS-, red, non-spotted
3 B-ss, black, spotted
1 bbss, red, spotted

This result occurs if the genes are separate genes and unlinked.  The ratios would be somewhat different if the genes were linked, but the four phenotypes would likely all show up.  So, the best answer is c.