Let $S_1$ be the subset of $S$ with all elements with remainder of $1$ when divided by $5$. $|S_1|=15$ Let $S_2$ be the subset of $S$ with all elements with remainder of $2$ when divided by $5$. $|S_2|=15$ Let $S_3$ be the subset of $S$ with all elements with remainder of $3$ when divided by $5$. $|S_3|=14$ Let $S_4$ be the subset of $S$ with all elements with remainder of $4$ when divided by $5$. $|S_4|=14$ Let $S_5$ be the subset of $S$ with all elements of multiple of $5$. $|S_5|=15$ Do case work with all possible combinations. #$1$, one and only one number of $A$ is from each of the subset of $S (S_1, S_2, S_3, S_4, S_5)$. The number of such subset $A$ is $${{15}\choose{1}} \cdot {{15}\choose{1}} \cdot {{14}\choose{1}} \cdot {{14}\choose{1}} \cdot {{15}\choose{1}}=661,500$$#$2$, two numbers of $A$ are from $S_1$, two numbers from $S_4$, and one number from $S_5$. The number of such subset $A$ is $${{15}\choose{2}} \cdot {{14}\choose{2}} \cdot {{15}\choose{1}}=143,325$$#$3$, two numbers of $A$ are from $S_2$, two numbers from $S_3$, and one number from $S_5$. The number of such subset $A$ is $${{15}\choose{2}} \cdot {{14}\choose{2}} \cdot {{15}\choose{1}}=143,325$$#$4$, All five numbers of $A$ are from one of the subset of $S (S_1, S_2, S_3, S_4, S_5)$. The number of such subset $A$ is $${{15}\choose{5}} + {{15}\choose{5}} + {{14}\choose{5}} + {{14}\choose{5}} + {{15}\choose{5}}=13,013$$Thus, the total number of the possible $A$ shall be $$661,500+143,325\times2+13,013=\boxed {961,163}$$

It is well known that the units digit of a perfect square can be $\{0, 1, 4, 5, 6, 9\}$. This would reduce the number of possible $m$. Definitely, the four-digit number can not have a leading digit of $0$. We have: \begin{align*} 1521 &=39^2, 1681=41^2 \\ 4764 &=68^2 \\ 5625 &=75^2 \\ 9409 &=97^2 \end{align*}There are total $\boxed{5}$ possible number $m \in \boxed{\{1521, 1681, 4764, 5625, 9409\}}$.

Let's choose $3$ male students and $3$ female students as candidates for the prizes. The number of combinations is $${{10} \choose {3}}\cdot {{12} \choose {3}}=26400$$Then we need to assign $6$ prizes to $6$ candidates. The number of possible assignations is $$\frac{6!}{1!2!3!}=60$$Then, number of possible arrangements is $$26400 \times 60 = \boxed{1,584,000}$$

Pairing the terms of RHS in the LHS so as to obtain $x^4-1$ and $x^2-9$ as denominators, leads to the inequality $$4x^2\left(\frac{1}{x^4-9}-\frac{1}{x^4-1}\right)\ge0$$

Maybe we could use cone area in this problem, Consider a point at AP as $X$ such that $BX$ is perpendicular to $AP$, Let the line be the radius of the base, by phytagoras since $AB=BP$ then $BX=12\sqrt{2}$,now consider point at AP as $Y$ such that $CY$ is perpendicular to $PA$, First we need to find $CP$,let $CP=x$ By Stewart $144(18)=324x+144(18-x)-(18-x)(x)18$ we get that $x=8$. Now by phytagoras again we could receive that $CY=\frac{16\sqrt{2}}{3}$. Then the volume will be $\frac{1}{3}(12\sqrt{2})^2(12)\pi-\frac{1}{3}(\frac{16\sqrt{2}}{3})^2(12)\pi=\frac{8320\pi}{9}$. I Don't know it is correct?