One. (20 points) There are three cylindrical containers $M$, $N$, and $P$, whose axial cross-sections are shown in figures 6(a), (b), and (c), respectively. The internal base areas are $S_{1} \mathrm{~cm}^{2}$, $S_{2} \mathrm{~cm}^{2}$, and $S_{3} \mathrm{~cm}^{2}$, and the internal heights are $h_{1} \mathrm{~cm}$, $h_{2} \mathrm{~cm}$, and $h_{3} \mathrm{~cm}$. The volumes are $V_{1} \mathrm{~cm}^{3}$, $V_{2} \mathrm{~cm}^{3}$, and $V_{3} \mathrm{~cm}^{3}$, respectively, with $V_{1} > V_{2} > V_{3}$. These three cylindrical containers $M$, $N$, and $P$ can be combined to form six different shapes of containers (by stacking the three containers from top to bottom in the order of $P N M$, $N P M$, $P M N$, $M P N$, $N M P$, and $M N P$), and their volumes are the sum of the volumes of $M$, $N$, and $P$.

Water is uniformly injected into these 6 containers at the same rate until they are full. The water depth $h(\mathrm{~cm})$ versus injection time $t(\mathrm{~s})$ for three of these containers is shown in figures 7, 8, and 9.

(1) Which of the six containers do the relationships in figures 7, 8, and 9 represent?
(2) Find the values of $h_{1}$, $h_{2}$, and $h_{3}$, and the ratio $S_{1}: S_{2}: S_{3}$.
(3) If $V_{3}=\frac{9}{m^{2}+4 m+4.125} \mathrm{~cm}^{3}$ and the injection rate is $\frac{n^{2}+6 n+45}{n+3} \mathrm{~cm}^{3} / \mathrm{s}$, where $m$ and $n$ are constants, find the sum of the volumes of containers $M$, $P$, and $N$.
(1) From Figure 7, we know that filling containers $M$, $N$, and $P$ takes $60 \mathrm{~s}$. From Figure 8, we know that filling two of the three containers $M$, $N$, and $P$ takes $54 \mathrm{~s}$. Therefore, filling the top container in Figure 8 takes $60-54=6(\mathrm{~s})$.

Similarly, filling the top container in Figure 9 takes $60-24=36(\mathrm{~s})$. Thus, filling the third container takes $60-6-36=18$ (s).

Since the water flow rate is constant and $V_{1}>V_{2}>V_{3}$, it follows that filling containers $M$, $N$, and $P$ takes $36 \mathrm{~s}$, $18 \mathrm{~s}$, and $6 \mathrm{~s}$, respectively.

Therefore, the patterns reflected in Figures 7, 8, and 9 are $N P M$, $P M N$, and $M N P$ respectively.
(2) Let the water flow rate be $V \mathrm{~cm}^{3} / \mathrm{s}$. From Figures 7, 8, 9, and the results of (1), we have:
$$
\left\{\begin{array}{l}
h_{1} + h_{3} = 24, \\
h_{2} + h_{1} = 30, \\
h_{2} + h_{3} = 18;
\end{array} \quad \left\{\begin{array}{l}
V_{1} + V_{2} + V_{3} = 60 V, \\
V_{2} + V_{1} = 54 V, \\
V_{3} + V_{2} = 24 V.
\end{array}\right.\right.
$$

Solving these equations, we get $h_{1}=18, h_{2}=12, h_{3}=6; V_{1}=36 \mathrm{~V}, V_{2}=18 \mathrm{~V}, V_{3}=6 \mathrm{~V}$.
Thus, $S_{1}=\frac{V_{1}}{h_{1}}=\frac{36 V}{18}=2 V$,
$$
\begin{array}{l}
S_{2}=\frac{V_{2}}{h_{2}}=\frac{18 V}{12}=\frac{3}{2} V, \\
S_{3}=\frac{V_{3}}{h_{3}}=\frac{6 V}{6}=V.
\end{array}
$$

Therefore, $S_{1}: S_{2}: S_{3}=2 V: \frac{3}{2} V: V=4: 3: 2$.
(3) Since $m^{2}+4 m+4.125=(m+2)^{2}+0.125$, the minimum value of $m^{2}+4 m+4.125$ is 0.125 when $m=-2$. Therefore, $V_{3} \leqslant 72$.
Given the water flow rate $V=\frac{n^{2}+6 n+45}{n+3}$, we have $n^{2}+(6-V) n+(45-3 V)=0$. Thus, $\Delta=(6-V)^{2}-4 \times 1 \times(45-3 V) \geqslant 0$.
Hence, $V \geqslant 12$ or $V \leqslant-12$ (discarded). When $n=3$, the minimum value of $V$ is 12.
From (2), we know $V_{3}=6 V$.
Since $V_{3} \leqslant 72$ and $6 V \geqslant 72$, we have $V_{3}=6 V=72$.
At this point, $m=-2, n=3$. Therefore, $V=12$.
Thus, $V_{1}+V_{2}+V_{3}=60 \mathrm{~V}=60 \times 12=720\left(\mathrm{~cm}^{3}\right)$.
Therefore, the total volume is $720 \mathrm{~cm}^{3}$.