Statement 1 is irrelevant.

Statement 2 is trickier than most GMAT DS statements. You might persuade yourself it's sufficient by picking a few numbers, but if you want to be sure of your answer, it's probably easiest to see why Statement 2 is sufficient by looking at numerical examples, and seeing how we can list all of the divisors of a number. Let's first take an odd number, say n = 45 = 3^2 * 5. n has six divisors, all of course odd:

1, 3, 5, (3^2), (3*5), (3^2 * 5) = 1, 3, 5, 9, 15, 45

Now, 2n will have all of those odd divisors, but will have just as many even divisors: you find the even divisors of 2n by doubling all of the odd divisors:

1, 3, 5, (3^2), (3*5), (3^2 * 5) = 1, 3, 5, 9, 15, 45
2*1, 2*3, 2*5, 2*(3^2), 2*(3*5), 2*(3^2 * 5) = 2, 6, 10, 18, 30, 90

are all of the divisors of 90. The same will be true of any odd n: if n is odd, 2n has twice as many divisors as n.

Instead start with an even number, say n = 54 = 2*(3^3). This number has four odd divisors, and four even divisors:

1, 3, 3^2, 3^3 = 1, 3, 9, 27
2, 2*3, 2*(3^2), 2*(3^3) = 2, 6, 18, 54

When we look at 2n = 108 = (2^2)*(3^3) , we'll have all of these divisors, but also all the divisors we get by doubling the divisors in the second row:

1, 3, 3^2, 3^3 = 1, 3, 9, 27
2, 2*3, 2*(3^2), 2*(3^3) = 2, 6, 18, 54
2^2, (2^2)*3, (2^2)*(3^2), (2^2)*(3^3) = 4, 12, 36, 108

We don't get twice as many, of course -- we get 50% more divisors. From this one example, hopefully it's clear why, for any even number n, 2n will never have twice as many divisors as n; 2n will always have less than twice as many if n is even.

Of course, if you understand why we find every divisor by doing as we did above, you'll likely know how, from the exponents in a prime factorization, to calculate the number of divisors of an integer. You can use that as well to answer the question.