Answer
See below
Work Step by Step
$A1$. Suppose $u=(x_1,y_1),v=(x_2,y_2) \in V=R^2$
Then $u+v=(x_1,y_1) \oplus (x_2,y_2)=(x_1-x_2,y_1-y_2) \in R^2$
Hence, $R^2$ is closed under addtion.
$A2$. Suppose $u=(x_1,y_1),v=(x_2,y_2) \in R^2, c \in R$
Then $c.u=c(x_1,y_1)=(-cx_1,-cy_1)$
Hence, $R^2$ is closed under scalar multiplication.
$A3$. Suppose $u=(x_1,y_1),v=(x_2,y_2) \in V$
Then $u \oplus v=(x_1,y_1) \oplus (x_2,y_2)=(x_1-x_2,y_1-y_2)\\v \oplus u=(x_2,y_2)\oplus (x_1,y_1)=(x_2-x_1) \oplus (y_2-y_1)$
$A4$. Suppose $u=(x_1,y_1),v=(x_2,y_2),z=(x_3,y_3) \in V$
then $(u \oplus v) \oplus z\\
=((x_1,y_1) \oplus (x_2,y_2)) \oplus (x_3,y_3)\\
=((x_1-x_2,y_1-y_2))\oplus (x_3-y_3)\\
=((x_1-x_2)-x_3, ((y_1-y_2)-y_3)\\
=(x_1-x_2-x_3,y_1-y_2-y_3)$
$u \oplus (v \oplus z)\\
=((x_1,y_1) \oplus (x_2,y_2)) \oplus (x_3,y_3)\\
=(x_1,y_1)\oplus (x_2-x_3,y_2-y_3)\\
=(x_1-(x_2-x_3), (y_1-(y_2-y_3))\\
=(x_1-x_2+x_3,y_1-y_2+y_3)$
Thus, $(u \oplus v) \oplus z \ne u \oplus (v \oplus z)$
$A5$. Suppose $u=(x_1,y_1) \in R^+$
then $u+0=(x_1,y_1)\oplus (0,0)=(x_1-0,y_1-0)=(x_1,y_1)$
Hence, $0$ is zero vector
$A6$. Suppose $u=(x_1,y_1) \in V$
then $u \oplus u=(x_1,y_1) \oplus (x_1,y_1)=(x_1-x_1,y_1-y_1)=(0,0)$
Hence, additive inverse of $(x_1,y_1)$ is also $u=(x_1,y_1)$.
$A7$. Suppose $u=(x_1,y_1) \in V$
then $-1 \oplus u=-1 \oplus (x_1,y_1)=((-1).x_1,-(-1).y_1)=(x_1,y_1)=u$
$A8$. Suppose $u=(x_1,y_1) \in V$ and $r,s \in R^+$
then $(rs) \oplus u\\
=(rs) \oplus (x_1,y_1)\\
=(-(rs)(x_1,-(rs)y_1))\\
=(-rsx_1,-rsy_1)$
and $r\odot (s \odot u)\\
=r \odot (-sx_1,-sy_1)\\
=(-(r(-sx_1),-r(-sy_1))\\
=(rsx_1,rsy_1)$
We can see that $(rs) \odot u \ne r \odot (s\odot u)$
$A9$. Suppose $u=(x_1,y_1),v=(x_2,y_2) \in V$ and $r \in R$
then $r \odot (u\oplus v)\\
=r \odot (x_1,y_1 \oplus x_2,y_2))\\
=r \odot (x_1-x_2, y_1-y_2)\\
=(-r(x_1-x_2),-r(y_1-y_2))\\
=(-rx_1+rx_2,-ry_1+ry_2)$
$r \odot u\oplus r \odot v\\
=r \odot (x_1,y_1) \oplus r \odot (x_2,y_2))\\
=(-rx_1,-rx_2)\oplus (rx_2, -ry_2)\\
=(-rx_1-(-rx_2),-ry_1-(-ry_2))\\
=(-rx_1+rx_2,-ry_1+ry_2)$
$A10$. Suppose $u=(x_1,y_1) \in V$ and $r,s \in R$
then $(r+s)\odot u\\
(r+s) \odot (x_1,y_1)\\
=(-(r+s)x_1,-(r+s)y_1)=(-rx_1-sx_1,-ry_1-sy_1)$
and $r \odot u \oplus s \odot u\\
=r \odot (x_1,y_1)\oplus s \odot (x_1,y_1)\\
=(-rx_1,-ry_1)\oplus (-sx_1,-sy_1)\\
=(-rx_1-(-sx_1),-ry_1-(-sy_1))\\
=(-rx_1+sx_1,-ry_1+sy_1)$
We can see that $(r+s)\oplus u \ne r \odot u \oplus s \odot u$
