(1) We perform the operation on $P_0$. It is easy to see that each side of $P_0$ becomes $4$ sides of $P_1$. So the number of sides of $P_1$ is $3 \cdot 4$. In the same way, we operate on $P_1$. Each side of $P_1$ becomes $4$ sides of $P_2$. So the number of sides of $P_2$ is $3 \cdot 4^2$. Consequently, it is not difficult to get that the number of sides of $P_n$ is $3 \cdot 4^n$.

It is known that the area of $P_0$ is $S_0 = 1$. Comparing $P_1$ with $P_0$, it is easy to see that we add to $P_1$ a smaller equilateral triangle with area $\frac{1}{3^2}$ on each side of $P_0$. Since $P_0$ has $3$ sides, so
$$S_1 = S_0 + 3 \cdot \frac{1}{3^2} = 1 + \frac{1}{3}.$$
Again, comparing $P_2$ with $P_1$, we see that $P_2$ has an additional smaller equilateral triangle with area $\frac{1}{3^2} \cdot \frac{1}{3^2}$ on each side of $P_1$, and $P_1$ has $3 \cdot 4$ sides. So that
$$S_2 = S_1 + 3 \cdot 4 \cdot \frac{1}{3^4} = 1 + \frac{1}{3} + \frac{4}{3^3}.$$
Similarly, we have
$$S_3 = S_2 + 3 \cdot 4 \cdot \frac{1}{3^6} = 1 + \frac{1}{3} + \frac{4}{3^3} + \frac{4^2}{3^5}.$$
Hence, we have
$$\begin{align*} S_n &= 1 + \frac{1}{3} + \frac{4}{3^3} + \frac{4^2}{3^5} + \dots + \frac{4^{n-1}}{3^{2n-1}} \\ &= 1 + \sum_{k=1}^{n} \frac{4^{k-1}}{3^{2k-1}} = 1 + \frac{3}{4} \sum_{k=1}^{n} \left(\frac{4}{9}\right)^k \\ &= 1 + \frac{3}{4} \cdot \frac{4}{9} \cdot \left[1 - \left(\frac{4}{9}\right)^n\right] \\ &= 1 + \frac{3}{5} \left[1 - \left(\frac{4}{9}\right)^n\right] \\ &= \frac{8}{5} - \frac{3}{5} \cdot \left(\frac{4}{9}\right)^n. \tag{*} \end{align*}$$

We will prove $(*)$ by mathematical induction as follows:
When $n=1$, it is known that $(*)$ holds from above.
Suppose, when $n=k$, we have $S_k = \frac{8}{5} - \frac{3}{5} \cdot \left(\frac{4}{9}\right)^k$.
When $n=k+1$, it is easy to see that, after $k+1$ times of operations, by comparing $P_{k+1}$ with $P_k$, we have added to $P_{k+1}$ a smaller equilateral triangle with area $\frac{1}{3^{2(k+1)}}$ on each side of $P_k$ and $P_k$ has $3 \cdot 4^k$ sides. So we get
$$\begin{aligned} S_{k+1} &= S_k + 3 \cdot 4^k \cdot \frac{1}{3^{2(k+1)}} \\ &= S_k + \frac{4^k}{3^{2k+1}} = \frac{8}{5} - \frac{3}{5} \cdot \left(\frac{4}{9}\right)^{k+1}. \end{aligned}$$
By mathematical induction, $(*)$ is proved.

(2) From (1), we have $S_n = \frac{8}{5} - \frac{3}{5} \cdot \left(\frac{4}{9}\right)^n$.
Therefore,
$$\lim_{n \to \infty} S_n = \lim_{n \to \infty} \left[ \frac{8}{5} - \frac{3}{5} \cdot \left(\frac{4}{9}\right)^n \right] = \frac{8}{5}.$$