= Solution
Let $i:X_{\mathrm{red}}\hookrightarrow X$ be the <reduced induced subscheme>, with <structure sheaf> $\mathcal O_{X_{\mathrm{red}}}=\mathcal O_X/\mathcal N$. Its underlying map of <topological spaces> is a <homeomorphism>, since every <prime ideal> contains the <nilradical>.
Put $\mathcal G=\mathcal F/\mathcal N\mathcal F$. Multiplication by $\mathcal N$ annihilates $\mathcal G$, so its action by $\mathcal O_X$ factors through $\mathcal O_X/\mathcal N$. Thus $\mathcal G$ is naturally a <sheaf of modules> over $\mathcal O_{X_{\mathrm{red}}}$. On an <affine open subscheme> $\operatorname{Spec}A$, with $\mathcal F=\widetilde M$, it corresponds to $M/\operatorname{Nil}(A)M$ over $A/\operatorname{Nil}(A)$, so it remains a <quasi-coherent sheaf>.
The <sheaf cohomology> of a <sheaf of modules> is the <sheaf cohomology> of its underlying <sheaf of abelian groups>. Under the above <homeomorphism>, that underlying sheaf is identical on the two spaces: the <global sections functor> and its <flasque resolutions> are identical. Equivalently, use <sheaf cohomology under a closed inclusion> with $i_*\mathcal G$. Therefore there are canonical <isomorphisms>, in every degree,
$$
\boxed{H^q(X,\mathcal F/\mathcal N\mathcal F)\cong H^q(X_{\mathrm{red}},\mathcal F/\mathcal N\mathcal F)\qquad(q\geq0).}
$$
In particular, this gives the requested equality in degree one. Equality of the underlying <topological spaces> alone would not compare arbitrary coefficient sheaves; the factorization of the action explains why the indicated coefficients are the same.
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