Lets begin with some notation. Let # mean not equal to and let > mean greater or equal to.
(1) can be rewritten in the following way n(n+1) # 3. It should be clear that n and n+1 are consecutive numbers and that any three consecutive numbers must contain at least one that has a factor of 3 in it. So, according to our statement both n-1 (and n+2) must be divisible by 3. It works for negative numbers and zero as well. Remember, zero is divisible by 3 since 0/3 = 0. And, neither n or n+1 can be zero since that would make their product divisible by 3. A is sufficient.
Lets look at (2). We can rewrite 3n+5 > k +8. Since k is a multiple of 3, lets substitute 3x for k where x is any non-negative integer, and by subtracting 5 from both sides: 3n > 3x + 3, which implies that n > x + 1. This tells us nothing about the divisibility of n-1. It merely tells us that n-1 is non-negative.
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