2II) Let $n$ be a positive integer greater than 7. $P$ is a convex polygon with $n$ sides. Any $n-3$ non-intersecting diagonals of $P$ divide $P$ into $n-2$ triangles. A triangle in the triangulation of $P$ is called an inner triangle if all its sides are diagonals of $P$. If the triangulation of $P$ has exactly two inner triangles, express the triangulation of $P$ in terms of $n$.
Consider the extraction of a partial convex polygon $A_{1} A_{2} \cdots A_{n}$ from $P$, where $A_{1} A_{2}, \cdots, A_{n-1} A_{n}$ are edges of $P$, and $A_{1} A_{n}$ is a diagonal of $P$. Consider all triangulations of this convex polygon, where the number of triangulations without any inner triangles of $P$ is denoted as $f(n)$, the number of triangulations with exactly one inner triangle of $P$ is denoted as $g(n)$, and the number of triangulations with exactly two inner triangles of $P$ is denoted as $h(n)$.

(1) It is easy to see that for $n \leqslant 2$, $f(n)=0$, and $f(3)=1$. When $n \geqslant 4$, consider the triangle $A_{1} A_{k} A_{n}$ in the triangulation that contains $A_{1} A_{n}$. Since $A_{1} A_{n}$ is a diagonal, either $A_{1} A_{k}$ or $A_{k} A_{n}$ must be an edge of $P$, so $k=2$ or $k=n-1$.

If $k=2$, the number of schemes is the number of triangulations of $A_{2} A_{3} \cdots A_{n}$ that do not contain any inner triangles, which is $f(n-1)$. The same applies when $k=n-1$, so
$$
f(n)=2 f(n-1)=\cdots 2^{n-3} f(3)=2^{n-3}, \quad n \geqslant 3.
$$

(2) It is easy to see that for $n \leqslant 4$, $g(n)=0$, and $g(5)=1$. When $n \geqslant 6$, consider the triangle $A_{1} A_{k} A_{n}$ in the triangulation that contains $A_{1} A_{n}$. If $k=2$ or $k=n-1$, the triangle $A_{1} A_{k} A_{n}$ is not an inner triangle. The remaining shapes can be triangulated in $g(n)$ ways each. If $k=3, \cdots, n-2$, the triangle $A_{1} A_{k} A_{n}$ is not an inner triangle. The triangulations of the shapes $A_{1} A_{2} \cdots A_{k}$ and $A_{k} A_{2} \cdots A_{n}$ that do not contain any inner triangles have $f(k) f(n-k+1)$ ways, so
$$
\begin{aligned}
g(n) & =2 g(n-1)+\sum_{k=3}^{n-2} f(k) f(n-k+1) \\
& =2 g(n-1)+(n-4) 2^{n-5},
\end{aligned}
$$

By rearranging, we get
$$
\begin{aligned}
g(n)-\binom{n-3}{2} 2^{n-5} & =2\left[g(n-1)-\binom{n-4}{2} 2^{n-6}\right]=\cdots \\
& =2^{n-5}\left[g(5)-\binom{2}{2} 2^{0}\right]=0.
\end{aligned}
$$

Thus, $g(n)=\binom{n-3}{2} 2^{n-5}, \quad n \geqslant 5$. Noting the properties of binomial coefficients, for $n \geqslant 3$, we always have $g(n)=\binom{n-3}{2} 2^{n-5}$.

(3) It is easy to see that for $n \leqslant 6$, $h(n)=0$, and $h(7)=2$. When $n \geqslant 8$, consider the triangle $A_{1} A_{k} A_{n}$ in the triangulation that contains $A_{1} A_{n}$.

If $k=2$ or $k=n-1$, the triangle $A_{1} A_{k} A_{n}$ is not an inner triangle. Considering the remaining shapes, there are $h(n-1)$ ways each. If $k=3, \cdots, n-2$, the triangle $A_{1} A_{k} A_{n}$ is an inner triangle. The shapes $A_{1} A_{2} \cdots A_{k}$ and $A_{k} A_{2} \cdots A_{n}$ must each contain exactly one inner triangle. So
$$
\begin{aligned}
h(n)= & 2 h(n-1)+\sum_{k=3}^{n-2}[g(k) f(n-k+1) \\
& + g(n-k+1)] \\
= & 2 h(n-1)+2 \sum_{k=3}^{n-2} g(k) f(n-k+1) \\
= & 2 h(n-1)+2 \sum_{k=3}^{n-2} \binom{k-3}{2} 2^{n-7} \\
= & 2 h(n-1)+2^{n-6} \binom{n-4}{3}. \\
& h(n)=\binom{n-3}{4} 2^{n-6}, \quad n \geqslant 7.
\end{aligned}
$$

By rearranging, we get
Returning to the original problem, taking all $(n-1)$-sided sub-shapes of $P$, we can obtain $n h(n-1)$ triangulations that contain exactly 2 inner triangles. Considering any triangulation that contains exactly 2 inner triangles, let it have $a$ triangles with two sides as edges of $P$, and $b$ triangles with one side as an edge of $P$, then
$$
\left\{\begin{array}{l}
a+b+2=n-2, \\
2 a+b=n,
\end{array}\right.
$$

Solving, we get $a=4$. Removing one of these 4 triangles from each triangulation, we get exactly one of the above $(n-1)$-sided triangulations, meaning each valid triangulation is counted 4 times in the above calculation. In summary, the number of triangulations is $\frac{n h(n-1)}{4}=n \binom{n-4}{4} 2^{n-9}$.