7. Given $S_{1}=2^{2}, S_{2}=2^{4}, \cdots, S_{n}=2^{2 n}$. Construct the rectangle $O A B C$ as follows: Mark points $O, A_{1}, A_{2}, \cdots, A_{n}$ on the $x$-axis; mark points $C_{1}, C_{2}, \cdots, C_{n}$ on the $y$-axis; and for any positive integer $k(1 \leqslant k \leqslant n)$, we have
$$
A_{k-1} A_{k}=x_{k}, C_{k-1} C_{k}=y_{k},
$$

where $A_{0}=C_{0}=O, A_{n}=A, C_{n}=C$, and $x_{k}, y_{k} \in \mathbf{Z}_{+}, x_{k} y_{k}=S_{k}$. Find:
(1) the maximum area of the rectangle $O A B C$ and the corresponding $\left(x_{1}, x_{2}, \cdots, x_{n}\right)$ and $\left(y_{1}, y_{2}, \cdots, y_{n}\right)$;
(2) the minimum area of the rectangle $O A B C$ and the corresponding $\left(x_{1}, x_{2}, \cdots, x_{n}\right)$ and $\left(y_{1}, y_{2}, \cdots, y_{n}\right)$.
7. (1) The maximum value of the rectangular area is
$$
\frac{1}{3}\left(4^{n}-1\right)\left(4^{n}+n-1\right) \text {, }
$$

At this time, $\left(x_{1}, x_{2}, \cdots, x_{n}\right)=\left(1,1, \cdots, 1,2^{2 n}\right)$,
$$
\left(y_{1}, y_{2}, \cdots, y_{n}\right)=\left(2^{2}, 2^{4}, \cdots, 2^{2 n-2}, 1\right) \text {; }
$$
(2) The minimum value of the rectangular area is $4\left(2^{n}-1\right)^{2}$,

At this time, $\left(x_{1}, x_{2}, \cdots, x_{n}\right)=\left(2^{2}, 2^{3}, \cdots, 2^{n+1}\right)$,
$\left(y_{1}, y_{2}, \cdots, y_{n}\right)=\left(2^{0}, 2^{1}, \cdots, 2^{n-1}\right)$

or $\left(x_{1}, x_{2}, \cdots, x_{n}\right)=\left(2^{0}, 2^{1}, \cdots, 2^{n-1}\right)$,
$\left(y_{1}, y_{2}, \cdots, y_{n}\right)=\left(2^{2}, 2^{3}, \cdots, 2^{n+1}\right)$

or $\left(x_{1}, x_{2}, \cdots, x_{n}\right)=\left(y_{1}, y_{2}, \cdots, y_{n}\right)$
$=\left(2^{1}, 2^{2}, \cdots, 2^{n}\right)$.
Since $x_{i} 、 y_{i}$ are positive integers, and $x_{i} y_{i}=S_{i}$, so

$\left\{x_{i}\right\} 、\left\{y_{i}\right\}$ are both finite sets.
Therefore, the area of the rectangle $O A B C$ must have a maximum and minimum value.
Lemma: The area of the rectangle $O A B C$ reaches its maximum value if and only if the following two conditions are met:
(i) For any positive integer $k(1 \leqslant k \leqslant n), x_{k}$ and $y_{k}$ are exactly one equal to 1;
(ii) $\left|\left(x_{1}+x_{2}+\cdots+x_{n}\right)-\left(y_{1}+y_{2}+\cdots+y_{n}\right)\right|$

reaches its minimum value.
Proof: When the area of the rectangle $O A B C$ reaches its maximum value, let the corresponding values be
$\left(x_{1}, x_{2}, \cdots, x_{n}\right),\left(y_{1}, y_{2}, \cdots, y_{n}\right)$.
(i) Necessity.
Assume there exists $k$ such that $x_{k} y_{k}$ are both greater than 1, and let
$x_{1}+x_{2}+\cdots+x_{n}=X, X-x_{k}=X_{k}$,
$y_{1}+y_{2}+\cdots+y_{n}=Y, Y-y_{k}=Y_{k}$.
Then the area of the rectangle $O A B C$ is
$S=X Y=\left(X_{k}+x_{k}\right)\left(Y_{k}+y_{k}\right)$.
Without loss of generality, assume $X_{k} \geqslant Y_{k}$.
Replace $x_{k} 、 y_{k}$ in $\left(x_{1}, x_{2}, \cdots, x_{n}\right) 、\left(y_{1}, y_{2}, \cdots, y_{n}\right)$ with $1 、 S_{k}$, then the area of the new rectangle is
$S^{\prime}=\left(X_{k}+1\right)\left(Y_{k}+S_{k}\right)$.
Thus, $S^{\prime}-S$
$=\left(X_{k}+1\right)\left(Y_{k}+S_{k}\right)-\left(X_{k}+x_{k}\right)\left(Y_{k}+y_{k}\right)$
$=\left(X_{k} y_{k}-Y_{k}\right)\left(x_{k}-1\right)>0$,

which contradicts that $S$ is the maximum value.
(ii) Sufficiency.
From (i), we have
$X+Y=S_{1}+S_{2}+\cdots+S_{n}+n \triangleq C$.
Then $S=X Y=X(C-X)$, which is a quadratic function of $X$ with a negative leading coefficient, and its axis of symmetry is $X=\frac{C}{2}$. Therefore, when $X$ is closest to $\frac{C}{2}$, the area $S$
reaches its maximum value. This is when $|X-Y|$ is minimized.
Thus, the area is maximized.
The lemma is proved.

First, solve problem (1).
From $2^{2 n}=1+2^{0}+2^{1}+\cdots+2^{2 n-1}$
$\Rightarrow 2^{2 n}>2^{2}+2^{4} \pi+\cdots \cdot+2^{2 n-2}$,
Therefore, when and only when
$$
\begin{array}{l}
\left(x_{1}, x_{2}, \cdots, x_{n}\right)=\left(1,1, \cdots, 1,2^{2 n}\right),, 47 \\
\left(y_{1}, y_{2}, \cdots, y_{n}\right)=\left(2^{2}, 2^{4}, \cdots, 2^{2 n-2}, 1\right)
\end{array}
$$

Conditions (i) and (ii) of the lemma are satisfied, and the maximum value of the area at this time is
$$
\frac{1}{3}\left(4^{n}-1\right)\left(4^{n}+n-1\right) \text {. }
$$

Next, solve problem (2), to find the minimum value of the area.
By the Cauchy-Schwarz inequality,
$$
\begin{array}{l}
\left(x_{1}+x_{2}+\cdots+x_{n}\right)\left(y_{1}+y_{2}+\cdots+y_{n}\right) \\
\geqslant\left(\sqrt{x_{1} y_{1}}+\sqrt{x_{2} y_{2}}+\cdots+\sqrt{x_{n} y_{n}}\right)^{2} \\
=\left(\sqrt{S_{1}}+\sqrt{S_{2}}+\cdots+\sqrt{S_{n}}\right)^{2} .
\end{array}
$$

If there exist $\left(x_{1}, x_{2}, \cdots, x_{n}\right) 、\left(y_{1}, y_{2}, \cdots, y_{n}\right)$ such that the equality in the above inequality holds, then the area can reach its minimum value.
By the equality condition of the Cauchy-Schwarz inequality,
$$
\frac{x_{1}}{y_{1}}=\frac{x_{2}}{y_{2}}=\cdots=\frac{x_{n}}{y_{n}},
$$

and $x_{1} y_{1}=S_{1}=4$, we know that $\left(x_{1}, y_{1}\right)$ has three possible values: $(4,1),(1,4),(2,2)$.
Thus, we get
$$
\begin{array}{l}
\left(x_{1}, x_{2}, \cdots, x_{n}\right)=\left(2^{2}, 2^{3}, \cdots, 2^{n+1}\right), \\
\left(y_{1}, y_{2}, \cdots, y_{n}\right)=\left(2^{0}, 2^{1}, \cdots, 2^{n-1}\right), \\
\text { or } \quad\left(x_{1}, x_{2}, \cdots, x_{n}\right)=\left(2^{0}, 2^{1}, \cdots, 2^{n-1}\right), \\
(\left.y_{1}, y_{2}, \cdots, y_{n}\right)=\left(2^{2}, 2^{3}, \cdots, 2^{n+1}\right) \\
\text { or } \quad\left(x_{1}, x_{2}, \cdots, x_{n}\right)=\left(y_{1}, y_{2}, \cdots, y_{n}\right) \\
=\left(2^{1}, 2^{2}, \cdots, 2^{n}\right),
\end{array}
$$

all of which satisfy the requirements, and the minimum value of the area is $4\left(2^{n}-1\right)^{2}$.