You're right - you shouldn't assume that there are only black and white socks (what a boring sock drawer that would be!). You have to see if the statements limit it to just those possibilities.

If the statements told us that there was exactly a 0.2 chance of a black sock and 0.8 chance of a white sock on the first pick, then we would easily know that those are the only two options. If two probabilities add to 1, they have to be the only two options (assuming that they are mutually exclusive, as black and white socks are. A sock can't be both).

Here, we're told "less than 0.2" in the first statement and "more than 0.8" in the 2nd, which might not seem to be enough at first glance. However, we're also told that there are exactly 8 socks. So we have to consider the possibilities:

1 black, 7 non-black --> a 1/8 or 0.125 chance of picking black on the first pick
2 black, 6 non-black --> a 2/8 or 0.25 chance of picking black

Here, we see that we're already greater than a 0.2 probability, so it must be that there is only one black sock. So we know for sure that the probability of 2 black socks = 0. Sufficient.

If we consider the 2nd statement, ignore the fact that it says white socks, and just think of those as "non-black":

7 non-black, 1 black --> 0.875 chance of a non-black

Again, this is the only case that would give us a probability greater than 0.8 for non-black socks, so we must have exactly 1 black sock. Sufficient.

The answer is D.