Groups acting on Riemann Surfaces By Hurwitz theorem, order of a group $G$ of automorphisms (conformal homeomorphisms) of a compact Riemann surface of genus $g\geq 2$ is bounded above by $84(g-1)$.
1. Is there any example of a compact Riemann surface whose automorphism group is trivial?
2. Does $C_2$ act on every compact Riemann surface of genus $g\geq 2$ ? ($C_2$ acts on any compact surface of genus $g$).
3. If all Sylow-subgroups of a finite group act on the a compact Riemann surface, does it imply that the whole group acts on Riemann surface?
4. Can one suggest a survey article about groups acting on Riemann surfaces/ automorphisms of Riemann surfaces? 
 A: A counterexample to Q3 is provided by the genus 2 compact Riemann surface $X$ of $y^2=x^5-1$. Indeed, the order 10 cyclic group $C_{10}$ acts on $X$ (by changing sign of $y$ and multiplying $x$ by $5$th roots of unity). It is known that the jacobian  of $X$ has endomorphism ring $Z[\zeta_5]$ - the $5$th cyclotomic ring of integers and any finite multiplicative subgroup of $Z[\zeta_5]$  is a subgroup of $\mu_{10}\cong C_{10}$. This implies that $Aut(X)=C_{10}$. On the other hand, the dihedral group $D_{10}$ of order $10$ has the same Sylow subgroups as $C_{10}$ but is not isomorphic to it. In other words, there is no faithful action of $D_{10}$ on $X$ while its Sylow subgroups $C_5$ and $C_2$ act faithfully on $X$.
If $Y$ is a compact Riemann surface of genus $g$ and its jacobian $J$ has no nontrivial automorphisms (i.e., $End(J)$ is the ring of integers $Z$) then either $Y$ is non-hyperelliptic and $Aut(Y)=\{1\}$ or $Y$ is hyperelliptic and $Aut(Y)=C_2$. For example, if $g>1$ and $Y_g$ is the hyperelliptic Riemann surface $y^2=x^{2g+1}-x-1$ then its jacobian $J_g$ has no nontrivial endomorphisms (Math. Research Letters 7 (2000), 123--132) and therefore $Aut(Y_g)=C_2$. If $p$ is an odd prime then for each integer $n \ge 5$ the automorphism group of the compact Riemann surface $y^p=x^n-x-1$ is the cyclic group $C_{p}$. Indeed, the endomorphism ring of the jacobian is the $p$th cyclotomic ring $Z[\zeta_p]$ (Math. Proc. Cambridge Philos. Soc.  136
(2004), 257--267) and one may easily check, using the differentials of the first kind that the curve is non-hyperelliptic.
Using Del Pezzo surfaces of degree 2, one may construct non-hyperelliptic genus 3 curves $Y$, whose jacobian has no nontrivial endomorphisms (AMS Translations Series 2, vol.  218 (2006), 67--75; MR2279305, 2007k:14060) and therefore $Aut(Y)=\{1\}$. For the genus 4 case see a paper of Anthony Várilly-Alvaradoa and David Zywina (LMS Journal of Computation and Mathematics (2009), 12: 144-165); their approach makes use of Del Pezzo surfaces of degree 1 (see also Math. Ann. 340 (2008),  407--435).
A: Q1: A typical Riemann surface has no holomorphic automorphisms, and this implies a negative answer to Q2. I don't see that Q3 can work: the p-subgroups surely don't uniquely determine the group in general. 
The literature on these questions is quite large. http://www.jstor.org/pss/2160738 is a paper on the issue of surfaces with no non-trivial automorphisms. It is a little hard to tell what you want, but some of the material on the inverse Galois theory problem (which does use curves) might help you.
A: I agree with the answers above.  Allen Broughton at Rose-Hulman is a guy who has written a lot about automorphisms of Riemann surfaces:  his paper Classifying finite group actions on surfaces of low genus, J. of Pure & Appl. Algebra 69 (1990), 233-270 will probably be of use.
