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$P=(x,y,p)$ implies $P\cap\mathbb Z=p\mathbb Z$ (here $p$ is a prime or $0$). As localization commutes with local cohomology $$H^2_{(x,y)}\left(\frac{\mathbb{Z}[x,y]}{(5x+4y)}\right)_P\simeq H^2_{(x,y)}\left(\frac{\mathbb{Z}[x,y]_P}{(5x+4y)}\right).$$ But $\mathbb Z[x,y]_P\simeq\mathbb Z_{(p)}[x,y]_{\overline{P}}$, where $\overline{P}$ is the extension of $P$ to $\mathbb Z_{(p)}[x,y]$. Now let's see what happens with the involved ideals via this isomorphism: $(5x+4y)$ goes to $(5x+4y)\mathbb Z_{(p)}[x,y]_{\overline{P}}$ and note that $5$ or $4$ (or both) are invertible in $\mathbb Z_{(p)}$, so the factor ring $\mathbb Z_{(p)}[x,y]_{\overline{P}}/(5x+4y)\mathbb Z_{(p)}[x,y]_{\overline{P}}$ is isomorphic to $\mathbb Z_{(p)}[t]_Q$, where $Q\cap \mathbb Z_{(p)}=p\mathbb Z_{(p)}$. Local cohomology is independent of base ring, so finally we arrive to $H^2_{(t)}(\mathbb Z_{(p)}[t]_Q)=0$ (since the local cohomology in a principal ideal is zero from $2$ onwards).