For part (2) , when 2 vectors are parallel to each other , one can be expressed as a scalar multiple of the other . IN this case ,

x=cy, where c is a scalar

(t+1)i+3j=ci+c(t-1)j

comparing , t+1=c ---1 , 3=c(t-1) --- 2 , solve them .

OR

Since they are parallel , their gradients are equal .

gradient of x = 3/(t+1)

gradient of y = t-1

set them equal , and solve for t .

(Bold is to indicate sth is a vector , else it would be a variable)