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Comment to the previous question: I don't think it would be possible to put restrictions on $X$ and $A$ such that $A^k\otimes L$ is ample for any line bundle $L$. If such a condition holds, then in particular every line bundle is ample (which is impossible unless $X$ is a point).

Perhaps by restricting the set of line bundles $L$ would give you better results, but even in this case it might require specific conditions on $X$, e.g., letting $L$ vary over the effective line bundles, would give that the closures of the ample cone and the effective cone are equal, which is a strong requirement.

A small remark for the 2nd question: How about something like this: Let $I=I_{p,q}$ be the ideal sheaf of $p\cup q$ and let $p:X'\to X$ be the blow-up of $X$ with center $I$ with exceptional divisor $E$. Let $A=O(1)$ and let $F$ be a coherent sheaf on $X'$. We want to show that $L=p^*O(n) \otimes O(-E)$ is ample, or equivalently that large powers of $L$ kill the higher cohomology of $F$. By the Leray spectral sequence,

$$ H^i(X',L^{\otimes k}\otimes F)=H^i(X, I^k(nk) \otimes p_* F). $$ This latter group vanishes for $i>0$ and $k/n$ large, since $A=O(1)$ is ample $(p_*O(−kE)=I^k$, all the higher direct images $R^i O(-kE)$ vanish, and $p_*F$ is coherent). So if we can show that this $n$ can be chosen independently of $p,q$ we should be done. Note that when $F=O_X$, $n$ can be chosen independently of $p,q$ since all $I$ have the same Hilbert polynomial.

I don't think it would be possible to put restrictions on $X$ and $A$ such that $A^k\otimes L$ is ample for any line bundle $L$. If such a condition holds, then in particular every line bundle is ample (which is impossible unless $X$ is a point).

Perhaps by restricting the set of line bundles $L$ would give you better results, but even in this case it might require specific conditions on $X$, e.g., letting $L$ vary over the effective line bundles, would give that the closures of the ample cone and the effective cone are equal, which is a strong requirement.

Comment to the 2nd question: How about something like this: Let $I=I_{p,q}$ be the ideal sheaf of $p\cup q$ and let $p:X'\to X$ be the blow-up of $X$ with center $I$ with exceptional divisor $E$. Let $A=O(1)$ and let $F$ be a coherent sheaf on $X'$. We want to show that $L=p^*O(n) \otimes O(-E)$ is ample, or equivalently that large powers of $L$ kill the higher cohomology of $F$. By the Leray spectral sequence,

$$ H^i(X',L^{\otimes k}\otimes F)=H^i(X, I^k(nk) \otimes p_* F). $$ This latter group vanishes for $i>0$ and $k/n$ large, since $A=O(1)$ is ample $(p_*O(−kE)=I^k$, all the higher direct images $R^i O(-kE)$ vanish, and $p_*F$ is coherent). So if we can show that this $n$ can be chosen independently of $p,q$ we should be done. Note that when $F=O_X$, $n$ can be chosen independently of $p,q$ since all $I$ have the same Hilbert polynomial.

How about something like this: Let $I=I_{p,q}$ be the ideal sheaf of $p$ and let $p:X'\to X$ be the blow-up of $X$ with center $I$ with exceptional divisor $E$. Let $A=O(1)$. Then by the Leray spectral sequence:$$H^i(X',p^*O(n) \otimes O(-E))=H^i(X, I(n))=0 $$for $n$ large enough since $A=O(1)$ is ample $(p_*O(−E)=I$ and all the higher direct images $R^i O(-E)$ vanish). This $n$ can be chosen independently of $p$ since all $I$ have the same Hilbert polynomial.

Comment to the previous question: I don't think it would be possible to put restrictions on $X$ and $A$ such that $A^k\otimes L$ is ample for any line bundle $L$. If such a condition holds, then in particular every line bundle is ample (which is impossible unless $X$ is a point).

Perhaps by restricting the set of line bundles $L$ would give you better results, but even in this case it might require specific conditions on $X$, e.g., letting $L$ vary over the effective line bundles, would give that the closures of the ample cone and the effective cone are equal, which is a strong requirement.

Comment to the previous question: I don't think it would be possible to put restrictions on $X$ and $A$ such that $A^k\otimes L$ is ample for any line bundle $L$. If such a condition holds, then in particular every line bundle is ample (which is impossible unless $X$ is a point).

Perhaps by restricting the set of line bundles $L$ would give you better results, but even in this case it might require specific conditions on $X$, e.g., letting $L$ vary over the effective line bundles, would give that the closures of the ample cone and the effective cone are equal, which is a strong requirement.

A small remark for the 2nd question: How about something like this: Let $I=I_{p,q}$ be the ideal sheaf of $p\cup q$ and let $p:X'\to X$ be the blow-up of $X$ with center $I$ with exceptional divisor $E$. Let $A=O(1)$ and let $F$ be a coherent sheaf on $X'$. We want to show that $L=p^*O(n) \otimes O(-E)$ is ample, or equivalently that large powers of $L$ kill the higher cohomology of $F$. By the Leray spectral sequence,

$$ H^i(X',L^{\otimes k}\otimes F)=H^i(X, I^k(nk) \otimes p_* F). $$ This latter group vanishes for $i>0$ and $k/n$ large, since $A=O(1)$ is ample $(p_*O(−kE)=I^k$, all the higher direct images $R^i O(-kE)$ vanish, and $p_*F$ is coherent). So if we can show that this $n$ can be chosen independently of $p,q$ we should be done. Note that when $F=O_X$, $n$ can be chosen independently of $p,q$ since all $I$ have the same Hilbert polynomial.

How about something like this: Let $I=I_{p,q}$ be the ideal sheaf of $p$ and let $p:X'\to X$ be the blow-up of $X$ with center $I$. Let $A=O(1)$. Then by the Leray spectral sequence $(p_*O(−E_p-E_q)=I$ and all the higher direct images $R^i O(-E_p-E_q)$ vanish): $$H^i(X',p^*O(n) \otimes O(-E_p))=H^i(X, I(n))=0 $$for $n$ large enough since $A=O(1)$ is ample. This $n$ can be chosen independently of $p$ since all $I$ have the same Hilbert polynomial.

Comment to the previous question: I don't think it would be possible to put restrictions on $X$ and $A$ such that $A^k\otimes L$ is ample for any line bundle $L$. If such a condition holds, then in particular every line bundle is ample (which is impossible unless $X$ is a point).

Perhaps by restricting the set of line bundles $L$ would give you better results, but even in this case it might require specific conditions on $X$, e.g., letting $L$ vary over the effective line bundles, would give that the closures of the ample cone and the effective cone are equal, which is a strong requirement.

How about something like this: Let $I=I_{p,q}$ be the ideal sheaf of $p$ and let $p:X'\to X$ be the blow-up of $X$ with center $I$ with exceptional divisor $E$. Let $A=O(1)$. Then by the Leray spectral sequence:$$H^i(X',p^*O(n) \otimes O(-E))=H^i(X, I(n))=0 $$for $n$ large enough since $A=O(1)$ is ample $(p_*O(−E)=I$ and all the higher direct images $R^i O(-E)$ vanish). This $n$ can be chosen independently of $p$ since all $I$ have the same Hilbert polynomial.

Comment to the previous question: I don't think it would be possible to put restrictions on $X$ and $A$ such that $A^k\otimes L$ is ample for any line bundle $L$. If such a condition holds, then in particular every line bundle is ample (which is impossible unless $X$ is a point).

Perhaps by restricting the set of line bundles $L$ would give you better results, but even in this case it might require specific conditions on $X$, e.g., letting $L$ vary over the effective line bundles, would give that the closures of the ample cone and the effective cone are equal, which is a strong requirement.