Let $R$ be a domain.
1) If $M$ is a non-zero injective $R$-module, then every non-zero homomorphism from $M$ to $R$ is surjective.
2) If $M$ is a non-zero projective $R$-module, then there exists a non-zero homomorphism from $M$ to $R$.
3) There exists a non-zero injective and projective $R$-module if and only if $R$ is a field.
My idea, since R is a domain and hence it has no zero-divisor, is to use the facts that injective implies divisible (i.e. multiplying by $r \neq 0$ is surjective) and that projective implies torsion-free (i.e. multiplying by $r \neq 0$ is injective) but I don't know how to continue. Any suggestions?
All of this assumes $M\neq \{0\}$ since all three break down in that case.
1) Good start for the first one. Suppose $M$ is injective and $f(m)=r\neq 0$. There exists $n$ such that $m=nr$. Then $r=f(m)=f(n)r$. Using cancellation, $1=f(n)$. Thus, $f$ is onto.
2) I couldn't immediately see a proof using the torsion-free idea you had. This is what occurred to me: since $M$ is projective, $M\oplus K\cong R^I$ for some index set $I$. This means there is an injection $j:M\to R^I$. Since $M$ is nonzero, $j$ is nonzero on some coordinate of the free module $R^I$. If $\pi$ is the projection onto that coordinate, then $\pi j:M\to R$ is nonzero.
3) By 2), there exists a nonzero homomorphism $f:M\to R$. By 1) it is surjective. Let $0\neq a\in R$, and $f(e)=1$. There exists $n\in M$ such that $na=e$, so $f(n)a=f(na)=f(e)=1$, and $a$ is invertible. Thus $R$ is a field.