We are given,
1.

2.

3.

a)
We have,
(using 3)

b)
Consider the identity

Now,
(using 3)
or
or(using 3 again)
or

c)
We have,


(using 3)
(using 3)


d)
It is merely a restatement of 2.()

Given only 1, 2, and 3, there is no mention of any "-1" element.
There are only 1, i, C, and various multiplications of these.

Now rule (1) allows you to cancel any 4 i's multiplied in a row;
Rule (2) allows you to cancel any 2 C's multiplied in a row;
and rule (3) allows you to take any 'string' of i's and C's, and move all the C's to the right, e.g.

CiiCi

= CiiiiiC (3)
= CiC (1)
= iiiCC (3)
= iii (2)

So, each element of the group can be represented as (0-3 i's)(0 or 1 C).
The 8 distinct elements are thus: 1, i, ii, iii, C, iC, iiC, and iiiC.

..of course there's no reason you can't then denote i^2 as '-1', but it makes no difference to the structure of the group.