For a number to be a multiple of 10, the unit digit should be 10, so \(t^2+1\) should have a 0 as units digit. So, \(t^2\) should have a 9 as units digit, meaning that t has to end in 3 or 7.

(1) \(91^2\) × t leaves a remainder of 1 when divided by 2.
t could be any odd number as 91^2*t is ODD.
If ending in 3 or 7 yes, otherwise no.
Insuff

(2) \(91^2\) × t = 5q+2, where q is a non-negative integer.
Now, RHS 5q+2 can end in two ways.
When q is even = 5*even+2=0+2=2, so it will end in 2 and \(91^2*t\) would also end in t, meaning that t has a units digit of 2, and then answer is NO.
When q is odd = 5*odd+2=5+2=7, so it will end in 7 and \(91^2*t\) would also end in t, meaning that t has a units digit of 7, and then answer is YES.
Insuff

Combined..
t is odd from I, so t cannot be ending in 2. t ends in 7 and so our answer is YES.

C