Let $C$ be a non-singular projective curve of genus $g\geq2$ over $\mathbb{C}$.
If $g\geq 3$ and $C$ is general, why does $C$ not admit etale coverings $\pi:C\rightarrow C'$ of degree $>1$? Is this due to the fact, that a general curve of genus $g\geq 3$ does not have automorphisms?
If $g=2$, why does no curve $C$ of this genus admit etale coverings $C\rightarrow C'$? (I think I figured this one out: this follows from the Riemann-Hurwitz formula.)
To answer the question about $C \to C'$, there are probably various ways to this, probably including easier ones than those I am going to suggest.
If you fix a genus $g',$ and consider etale covers of some fixed degree $d$ of $C'$, you get curves $C$ of some prescribed genus $g$ (given by the Riemann--Hurwitz formula). Now $C'$ depends on $3g' - 3$ parameters (this is the dimension of the moduli space of genus $g'$ curves), and it has only finitely many etale covers of fixed degree. So the curves $C$ which admits etale morphisms $C \to C'$ depend only on $3g'-3$ parameters. But the moduli space of genus $g$ curves is of dimension $3g -3$, so most curves do not arise as such etale covers.
Another way to think about the problem (but I don't know how to actually prove it this way) is that if we have an etale cover $C \to C'$ then we obtain a morphism of Jacobians $\mathrm{Jac}(C') \to \mathrm{Jac}(C)$, and so $\mathrm{Jac}(C)$ is not a simple abelian variety. But, heuristically, there is no reason to expect that for a general curve $C$, its Jacobian wouldn't be simple. (One problem with this approach is that most abelian varieties don't arise as Jacobians at all, and so, while I find it suggestive, it isn't more than that, as it stands.)