On the algebraic fundamental group of surfaces with K^2\leq 3\chi

Let S be a minimal complex surface of general type with $q(S)=0$. We prove the following statements concerning the algebraic fundamental group: I) Assume that K^2_S\leq 3\chi(S). Then S has an irregular etale cover if and only if S has a free pencil of hyperelliptic curves of genus 3 with at least 4 double fibres. II) If K^2_S=3 and \chi(S)=1, then S has no irregular etale cover. III) If K^2_S<3\chi(S) and S does not have any irregular etale cover, then the order of the algebraic fundamental group is lesser or equal to 9, and if equality occurs then K^2_S=2, \chi(S)=1.