Regard $K=\mathbb{R}-\lbrace{0\rbrace}$ as a multiplication group. Let $f:K\to K$ be a multiplication homormorphism.


Question 1. Whether that $f$ is surjective implies that $f$ is injective?


Question 2. Whether that $f$ is injective implies that $f$ is surjective?


Question 3. $g: x\to x^b$ is a multiplication homormorphism of $K$ where $b=n/m, (n,m)=1$,$n\in\mathbb{Z}$ and $m$ is an odd integer. How to find any other multiplication homormorphism of $K$ than this form. Any example? 


Edit. Emil Jeřábek gave other explicit examples:
$h: x\to |x|^r$ or $x\to sgn(x)|x|^r.$ Of course, $hg$ is also ok.


Any other explicit ones? 


Perhaps, these questions look like homework, but not easy to me to answer (my major is not in algebraic theory).



Questions about multiplicative homomorphism of $\mathbb{R}$
There is an isomorphism $\mathbb{Z}/2\mathbb{Z}\times (\mathbb{R}^+,\cdot )\to K$ given by $(x,y)\mapsto (-1)^xy$ (considering $\mathbb{Z}/2\mathbb{Z}$ to contain 0 and 1).
There is also an isomorphism $(\mathbb R,+)\to(\mathbb R^+,\cdot)$ given by $x\mapsto exp(x)$.


So for all your questions it is enough to consider $K':= (\mathbb R,+)$, since $K \cong \mathbb{Z}/2\mathbb{Z} \times K'$
But $K'$ is a $\mathbb{Q}$-vector space. So you can pick any basis (which will be uncountable) and then do something on it giving you a lot of endomorphisms of $K'$.


Both questions 1 and 2 are false, since on the uncountable basis of $K'$ there are injective but not surjective set-maps (and vice versa), that extend to an endomorphism of $K'$.


