Actually, you have done extremely well!

When I did the problem, I got (1, 0, -1, 0) and (0, 1, 0, -1) but since those are just multiples of your vectors, they are both bases. Yes, your answer is correct.

I assume that your is the field of integers modulo 3 so that 2= -1 and the two equations we get from applying the matrix to any member, , of the kernel, x+ 2y+ z+ 2w= 0 and x- y+ z- w= 0, are exactly the same. Since there is only one condition, the dimension of the kernel is 4- 1= 3 rather than 4- 2= 2 as before. We can solve the second equation for x: x= y- z+ w and so .

Since, in , 2= -1, that is exactly what you have.