= Solution
A <positive holomorphic line bundle> $F$ admits a <Hermitian metric> whose <Chern connection> curvature satisfies that $iF_F$ is a <positive real (1, 1)-form>. In a local <holomorphic local frame> with squared length $e^{-\varphi}$, the <local formula for the Chern connection on a line bundle> gives $F_F=\partial\bar\partial\varphi$, so positivity means $i\partial\bar\partial\varphi$ is positive definite. Its closedness makes $\omega=iF_F$ a <Kähler form>; use this form to define the operators below.
Let $n=\dim_{\mathbb C}X>0$, choose a <Hermitian metric> on $E$, and equip $G_m=E\otimes F^{-m}$ with the tensor-product metric. The <curvature of a tensor product connection> gives
$$
iF_{G_m}=iF_E-m\omega\operatorname{id}_E.
$$
On $G_m$-valued zero-forms, the <Lefschetz commutator> is $[L,\Lambda]=-n\operatorname{id}$. Thus the <Bochner-Kodaira-Nakano identity> gives
$$
\Delta''_{G_m}=\Delta'_{G_m}+R_E+mn\operatorname{id},
\qquad R_E=[iF_E,\Lambda]\big|_{\mathcal A^{0,0}(E)}.
$$
This last operator is a fixed smooth self-adjoint bundle endomorphism. Compactness supplies a finite $C$ such that $\langle R_Ev,v\rangle\geq-C\|v\|^2$ at every point. For a <holomorphic section> $s$ of $G_m$, $D''s=0$ and $D''^*s=0$ by degree, so integrating the identity gives
$$
0=\|D's\|^2+\langle R_Es,s\rangle+mn\|s\|^2
\geq(mn-C)\|s\|^2.
$$
Choose an integer $m_0$ with $m_0n>C$. Then \b[$\boxed{H^0(X,E\otimes F^{-m})=0\quad(m\geq m_0)}$]. The threshold depends on the fixed bundle $E$, as the curvature bound makes explicit. Positive complex dimension is necessary: on a zero-dimensional manifold positivity is vacuous and a nonzero fibre has nonzero sections for every twist.
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