12. Starting from the equilateral triangle shown in Figure 4 (a), divide each side into three equal segments, and draw a small equilateral triangle outward on the middle segment of each side, forming the six-pointed star shape shown in Figure 4 (b); then, on each side of the six-pointed star, use the same method to draw even smaller equilateral triangles outward, forming a shape with 18 points as shown in Figure 4 (c); then, on each side of this shape, use the same method to draw even smaller equilateral triangles outward (as shown in Figure 4 (d)). Continue this process, and the outline of the shape will form a curve with increasingly more branches, which is the Koch snowflake curve idealized by Swedish mathematician Koch.
Figure 4
If the side length of the original equilateral triangle is $a$, let's call each step of the shape transformation a "growth". For example, after the first growth, we get Figure 4 (b), where each small equilateral triangle has a side length of $\frac{1}{3} a$, and the perimeter of the formed shape is $4 a$. Please fill in Table 1 (expressed in algebraic terms involving $a$).
Table 1
\begin{tabular}{|l|c|c|c|c|c|}
\hline & \begin{tabular}{l} 
First \\
Growth
\end{tabular} & \begin{tabular}{l} 
Second \\
Growth
\end{tabular} & \begin{tabular}{l} 
Third \\
Growth
\end{tabular} & $\cdots$ & \begin{tabular}{l} 
Growth
\end{tabular} \\
\hline Each small equilateral \\
triangle's side length & $\frac{1}{3} a$ & & & $\cdots$ & \\
\hline \begin{tabular}{c} 
Perimeter of the \\
formed shape
\end{tabular} & $4 a$ & & & $\cdots$ & \\
\hline
\end{tabular}
12. As shown in Table 2.
Table 2
\begin{tabular}{|c|c|c|c|c|c|}
\hline & \begin{tabular}{l} 
First \\ 
Growth
\end{tabular} & \begin{tabular}{l} 
Second \\ 
Growth
\end{tabular} & \begin{tabular}{l} 
Third \\ 
Growth
\end{tabular} & $\cdots$ & \begin{tabular}{l} 
$n$-th \\ 
Growth
\end{tabular} \\
\hline \begin{tabular}{l}
\begin{tabular}{c} 
Side Length \\ 
of Each \\ 
Small Equilateral \\ 
Triangle
\end{tabular}
\end{tabular} & \begin{tabular}{l}
$\frac{1}{3} a$
\end{tabular} & $\frac{1}{9} a$ & $\frac{1}{27} a$ & $\cdots$ & \begin{tabular}{l}
$\left(\frac{1}{3}\right)^{n} a$
\end{tabular} \\
\hline \begin{tabular}{l}
\begin{tabular}{l} 
Perimeter of \\ 
the Formed \\ 
Shape
\end{tabular}
\end{tabular} & $4 a$ & \begin{tabular}{l}
$\frac{16}{3} a$
\end{tabular} & \begin{tabular}{l}
$\frac{64}{9} a$
\end{tabular} & $\cdots$ & $\frac{4^{n}}{3^{n-1}} a$ \\
\hline
\end{tabular}