Answer
See below
Work Step by Step
$A1$. Suppose $x,y \in R^+$
Then $x \oplus y=x.y \in R^+$
Hence, $R^+$ is closed under addtion.
$A2$. Suppose $x\in R^+, c \in R$
Then $x \odot y=x^c \in R^+$
Hence, $R^+$ is closed under scalar multiplication.
$A3$. Suppose $x,y \in R^+$
Then $x \oplus y=x.y=y.x=y \oplus x$
$A4$. Suppose $x,y,z \in R^+$
then $(x \oplus y) \oplus z=(x.y).z=x.(y.z)=x \oplus (y \oplus z)$
$A5$. Suppose $x \in R^+$
then $x \oplus 1=x.1=x$
Hence, $1$ is zero vector
$A6$. Suppose $x \in R^+$
then $x \oplus \frac{1}{x}=x.\frac{1}{x}=1$
Hence, additive inverse of $x$ is $\frac{1}{x}$.
$A7$. Suppose $x \in R^+$
then $1 \oplus x=x^1=x$
$A8$. Suppose $x \in R^+$ and $r,s \in R^+$
then $(rx) \oplus x=x^{rx}=x^{r(x)}=r \oplus (x^s)=r \odot (s \odot x)$
$A9$. Suppose $x,y \in R^+$ adn $r \in R$
then $r \odot (x \oplus y)=(x \oplus y)^r=x^r.y^r=r \odot x \oplus r \odot y$
$A10$. Suppose $x \in R^+$ and $r,s \in R$
then $(r+s)\odot x=x^{r+s}=x^r.x^s=r \odot x \oplus s \odot x$
