Answer:
See below
Step-by-step explanation:
I will denote your new addition by +', and your new scalar multiplication by * to distinguish them from the usual operations.
1) x+'y=y+'x. Yes: Let x=[a,b] and y=[c,d]. Then x+'y=[0,b+d]=[0,d+b]=y+'x
2) (x+'y)+'z=x+'(y+'z). Yes: Let x=[a,b], y=[c,d], z=[e,f]
Then (x+'y)+'z=[0,b+d]+'[e,f]=[0,(b+d)+f]=[0,b+(d+f)]=[a,b]+'[0,d+f]=x+'(y+'z)
3) There exists an element 0 in V such that x+'0=x for each x in V: No
Take x=[1,0]. Then x+'[a,b]=[0,b]≠[1,0], so 0=[a,b] does not exist (no choice of 0 works for this x)
4) For each x in V, there exists an element -x in V such that x+'(-x)=0. Yes (if 0=[0,0])
Given x=[a,b], take -x=[a,-b]. Then x+'(-x)=[0,b-b]=[0,0]=0.
5) *k(x+y)=(*kx)+'(*ky) for each scalar k and any x and y in V. Yes
Let x=[a,b] and y=[c,d]. Then *k(x+'y)=*k[0,b+d]=[0,k(b+d)]=[0,kb]+[0,kd]=*kx+*ky.
6) (s+t)*x=(*sx)+(*tx) for any scalars s and t and any x in V. Yes
Let x=[a,b]. Then (s+t)*x=[0,(s+t)b]=[0,sb]+[0,tb]=*sx+*tx
7) (*s)(*tx)=*(st)x for any scalars s and t and any x in V. Yes
Let x=[a,b]. Then (*s)(*tx)=*s[0,tb]=[0,stb]=*(st)x
8) *1x=x for all x in V. No
*1[1,2]=[0,2]≠[1,2]
