= Solution
Suppose, to the contrary, that $H^1(X,\mathcal F)=0$ for every <quasi-coherent sheaf> $\mathcal F$. Since $X$ is a <quasi-compact scheme>, part (a) makes it an <affine scheme>.
Every <affine scheme> $X=\operatorname{Spec}A$ is a <separated scheme>: its <diagonal morphism> over $\operatorname{Spec}\mathbb Z$ is induced by the surjective <ring homomorphism> $A\otimes_{\mathbb Z}A\to A$, $a\otimes b\mapsto ab$, so it is a <closed immersion>. This contradicts the assumption that $X$ is not a <separated scheme>. Therefore
$$
\boxed{\text{There exists a quasi-coherent }\mathcal F\text{ with }H^1(X,\mathcal F)\ne0.}
$$
In fact, the version of the <Serre criterion for affineness> using ideals shows that $\mathcal F$ may be chosen to be a <quasi-coherent sheaf> of ideals. This argument does not use a comparison with <Čech cohomology> on the nonseparated <scheme>.
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