Data Sufficiency

t_0 = 3
t_1 = 3 + 1
t_2 = 3 + 1 + 2
t_3 = 3 + 1 +2 + 3
t_4 = 3 + 1 + 2 + 3 + 4

and so on, so t_n is just equal to 3 plus the sum of the first n positive integers. The sum of the first n positive integers is just (n)(n+1)/2 (it's the number of terms, n, times the average term, which, since it's an equally spaced list, is the average of the first and last terms, so is (n+1)/2 ). So we know:

t_n = 3 + (n)(n+1)/2

So t_n will be even only when (n)(n+1)/2 is even.

Statement 1 alone isn't much help - n might be 5 or 8, and then (n)(n+1)/2 will be odd or even, respectively. But Statement 2 is sufficient, because if n-1 is divisible by 4, then n is odd, and n+1 is only divisible by a single 2, so (n+1)/2 is odd. Since n and (n+1)/2 are both odd, so is (n) * (n+1)/2. So 3 + (n)(n+1)/2 will necessarily be a sum of two odd numbers, and thus is even.