I3.1 Given that $a$ is a positive real root of the equation $2^{x+1}=8^{\frac{1}{x}-\frac{1}{3}}$. Find the value of $a$.
I3.2 The largest area of the rectangle with perimeter $a$ meter is $b$ square meter, find the value of $b$.
I3.3 If $c=\left(1234^{3}-1232 \times\left(1234^{2}+2472\right)\right) \times b$, find the value of $c$.
I3.4 If $\frac{1}{(c+1)(c+2)}+\frac{1}{(c+2)(c+3)}+\cdots+\frac{1}{(c+d)(c+d+1)}=\frac{8}{15}$, find the value of $d$.
$$
\begin{array}{l}
2^{x+1}=8^{\frac{1}{x}-\frac{1}{3}} \Rightarrow 2^{x+1}=2^{\frac{3}{x}-1} \Rightarrow x+1=\frac{3}{x}-1 \\
x^{2}+2 x-3=0 \\
(x+3)(x-1)=0 \\
x=-3 \text { or } 1, a=1 \text { is a real positive root. }
\end{array}
$$
$x=-3$ or $1, a=1$ is a real positive root.
The perimeter $=1 \mathrm{~m}$.
Let the length of the rectangle be $x \mathrm{~m}$, the width is $\frac{1}{2}(1-2 x) \mathrm{m}$.
Its area is $\frac{1}{2}(1-2 x) \cdot x \mathrm{~m}^{2}=-\left(x^{2}-\frac{1}{2} x\right) \mathrm{m}^{2}=\left[-\left(x-\frac{1}{4}\right)^{2}+\frac{1}{16}\right] \mathrm{m}^{2}$.
$$
b=\frac{1}{16}=0.0625
$$

Remark: The original version is: The area of the largest rectangle ...
It is ambiguous to define the largest rectangle. It should be changed to "The largest area of the rectangle .....",
$$
\begin{array}{l}
c=\left(1234^{3}-1232 \times\left(1234^{2}+2472\right)\right) \times \frac{1}{16}, \text { let } x=1234 \\
=\frac{1}{16}\left\{x^{3}-(x-2) \times\left[x^{2}+2(x+2)\right]\right\}=\frac{1}{16}\left\{x^{3}-(x-2) \times\left(x^{2}+2 x+4\right)\right\} \\
=\frac{1}{16}\left\{x^{3}-\left(x^{3}-8\right)\right\}=\frac{8}{16}=\frac{1}{2} \\
\frac{1}{(c+1)(c+2)}+\frac{1}{(c+2)(c+3)}+\cdots+\frac{1}{(c+d)(c+d+1)}=\frac{8}{15} \\
\left(\frac{1}{c+1}-\frac{1}{c+2}\right)+\left(\frac{1}{c+2}-\frac{1}{c+3}\right)+\cdots+\left(\frac{1}{c+d}-\frac{1}{c+d+1}\right)=\frac{8}{15} \\
\frac{1}{c+1}-\frac{1}{c+d+1}=\frac{8}{15} \Rightarrow \frac{1}{\frac{1}{2}+1}-\frac{1}{\frac{1}{2}+d+1}=\frac{8}{15} \\
\frac{2}{3}-\frac{8}{15}=\frac{2}{3+2 d} \\
\frac{2}{15}=\frac{2}{3+2 d} \\
\frac{3}{d}=2 d=6
\end{array}
$$