List 5 $A$ produces gasoline, $B$ needs gasoline. Cars transport gasoline from $A$ to $B$, and the fuel consumption for a round trip is exactly equal to the tonnage of gasoline when the car is fully loaded, so it is impossible to transport gasoline directly from $A$ to $B$. A fuel depot is set up at an intermediate station $C$ on the way. First, cars traveling between $A$ and $C$ transport gasoline to $C$, and then cars traveling between $C$ and $B$ transport gasoline to $B$.
(1) If $A C=\frac{1}{3} A B$, how should the transportation be organized to transport gasoline from $A$ to $B$ in the most economical way (maximizing the amount of gasoline received at $B$)? What is the fuel delivery rate $\left(\frac{\text{amount of gasoline received at } B}{\text{amount of gasoline transported from } A}\right)$ in this case?
(2) If $A C=d$, what value of $d$ maximizes the fuel delivery rate? How should the transportation be organized in this case?
(3) If two intermediate stations $C$ and $D$ are set up, with $A C=d_{1}$ and $C D=d_{2}$, what values of $d_{1}$ and $d_{2}$ maximize the fuel delivery rate? How should the transportation be organized in this case?
Let $AB = l$, the number of cars shuttling between $A$ and $C$ be $x$, the number of cars shuttling between $C$ and $B$ be $y$, and the full load of gasoline per car be $a$.
(1) The amount of gasoline that can be transported out from station $C$ is $ax - \frac{ax}{3} = \frac{2ax}{3}$, and the amount of gasoline input to location $B$ is $ay - \frac{2ay}{3} = \frac{ay}{3}$. To ensure economic efficiency, each car should be fully loaded, i.e., $\frac{2ax}{3} = ya$, so $\frac{y}{x} = \frac{2}{3}$. If a fleet consists of $5m$ cars, with $3m$ cars shuttling between $A$ and $C$ and $2m$ cars shuttling between $C$ and $B$, each car can be fully loaded, maximizing the amount of gasoline received at $B$. In this case, the oil transportation rate $P = \frac{\frac{ay}{3}}{ax} = \frac{2}{9}$.
(2) The amount of gasoline that can be transported out from station $C$ is $ax - \frac{d}{l}ax = \frac{l-d}{l}ax$, and the amount of gasoline input to location $B$ is $ay - \frac{l-d}{l}ay = \frac{d}{l}ay$. To ensure $\frac{l-d}{l}ax = ay$, we have $\frac{y}{x} = \frac{l-d}{l}$, so the oil transportation rate $P = \frac{\frac{d}{l}ay}{ax} = \frac{d}{l} \cdot \frac{l-d}{l} = \frac{d(1-d)}{l^2} \leq \frac{1}{l^2}\left(\frac{d + l - d}{2}\right)^2 = \frac{1}{4}$. When $d = l - d$, i.e., $d = \frac{l}{2}$, $P_{\text{max}} = \frac{1}{4}$, and $\frac{y}{x} = \frac{1}{2}$. Therefore, when $d = \frac{l}{2}$, i.e., $C$ is located at the midpoint of $AB$, a fleet of $3m$ cars, with $2m$ cars shuttling between $A$ and $C$ and $m$ cars shuttling between $C$ and $B$, achieves the maximum oil transportation rate of $\frac{1}{4}$.
(3) Let the number of cars shuttling between $A$ and $C$, $C$ and $D$, and $D$ and $B$ be $x$, $y$, and $z$ respectively. The amount of gasoline that can be transported out from station $C$ is $ax - \frac{d_1}{l}ax = \frac{l-d_1}{l}ax$, and the amount of gasoline that can be transported out from station $D$ is $ay - \frac{d_2}{l}ay = \frac{l-d_2}{l}ay$. The amount of gasoline input to location $B$ is $az - \frac{l-d_1-d_2}{l}az = \frac{d_1 + d_2}{l}az$. To ensure $\frac{l-d_1}{l}ax = ay$ and $\frac{l-d_2}{l}ay = az$, we have $\frac{y}{x} = \frac{l-d_1}{l}$, $\frac{z}{y} = \frac{l-d_2}{l}$, and $\frac{z}{x} = \frac{(l-d_1)(l-d_2)}{l^2}$. Thus, the oil transportation rate $P = \frac{\frac{d_1 + d_2}{l}az}{ax} = \frac{(d_1 + d_2)(l-d_1)(l-d_2)}{l^3} \leq \frac{1}{l^3}\left(\frac{d_1 + d_2 + l - d_1 + l - d_2}{3}\right)^3 = \frac{8}{27}$. Therefore, when $d_1 + d_2 = l - d_1 = l - d_2$, i.e., $d_1 = d_2 = \frac{l}{3}$, $P_{\text{max}} = \frac{8}{27}$. At this point, $\frac{y}{x} = \frac{2}{3}$ and $\frac{z}{x} = \frac{4}{9}$, so $x : y : z = 9 : 6 : 4$. Hence, when $d_1 = d_2 = \frac{l}{3}$, i.e., $C$ and $D$ are located at the two trisection points of $AB$, a fleet of $19m$ cars, with $9m$ cars shuttling between $A$ and $C$, $6m$ cars shuttling between $C$ and $D$, and $4m$ cars shuttling between $D$ and $B$, achieves the maximum oil transportation rate of $\frac{8}{27}$.