Let $X$ be a smooth complex projective variety of dimension $n$, and let $\mathcal{F}$ be a globally generated rank $n$ vector bundle on $X$. Let $D$ be a smooth divisor on $X$. 


Is it true that there is a dense subset $U\subset H^0(\mathcal{F})$ such that for all $s\in U$, the zero set of $s$ does not intersect $D$?



Set of sections whose zeroes avoid a given divisor is (Zariski) dense?
Yes. Consider the incidence variety $Z\subset D\times \mathbb{P}(H^0(E))$ of pairs $(x,[s])$ with $x\in D$, $s\in H^0(E)\smallsetminus \{0\} $ and $s(x)=0$. Let $p,q$ be the projections from $Z$ to $D$ and $\mathbb{P}(H^0(E))$. For $x\in D$, the fiber $p^{-1}(x)$ is the subspace of $\mathbb{P}(H^0(E))$ formed by sections vanishing at $x$; since $E$ is globally generated, these fibers have codimension $n$, hence $Z$ has dimension $h^0(E)-2$. Therefore $q(Z)$ is a strict closed subvariety of $\mathbb{P}(H^0(E))$; any section in its complement does not vanish at any point of $D$. 


