A few questions about vector bundles on an algebraic variety Let $X$ be a smooth projective complex variety and $E$ an algebraic vector bundle on $X$.
(Q1) If $E$ is globally generated and $c_1(E)=0$ does it follows that $E$ is trivial?
(Q2) If $E$ is globally generated and $Q$ is a quotient of $E$, is $Q$ also globally generated?
(Q3) If $E$ is globally generated, $Q$ is a quotient of $E$ and $H^0(E^\ast)=0$, then must $Q$ be non-trivial?
 A: For 1 and 2, the answer is yes, depending on what you mean by $c_1$ of course. In 3, I do not know what $F$ is.
2 is trivial, since globally generated just means there exists a surjection $G\to E$, where $G$ is trivial. Then, obviously taking composite you get a surjection $G\to Q$.
1 is not difficult either. If rank of $E$ is $r$ and $E$ is globally generated, then taking $r$ general sections you can get an inclusion $G\to E$ where $G$ is a trivial bundle of rank $r$. Taking determinants, you get an inclusion $\Lambda^r G\to \Lambda^r E$. So, if you mean by $c_1(E)=0$ to mean that $\Lambda^rE=\mathcal{O}_X$, then the above inclusion has to be an isomorphism and thus the original map $G\to E$ must also be an isomorphism.
Of course, most definitions would imply that a globally generated line bundle with $c_1=0$ is just $\mathcal{O}_X$.
With your edit, in 3, if $Q\neq 0$ and trivial, then so is $Q^∗$ and it is a subsheaf of $E^∗$ and so $H^0(E^∗)\neq 0$. I assume when you say trivial, it is isomorphic to $\mathcal{O}_X^r$ for $r$ the rank. If $Q$ is trivial, so is $Q^*$ and non-zero means it has non-zero sections.
