12. The chessboard is marked with stations 0, 1, 2, ..., 100. The chess piece starts at station 0. The player tosses a fair coin to play the game. If heads come up, the piece moves forward one station; if tails come up, the piece moves forward two stations. The game continues until the piece reaches station 99 (Victory Camp) or station 100 (Failure Camp), at which point the game ends. Let the probability of the piece landing on station $n$ be $P_{n}$.
(1) Find the value of $P_{3}$;
(2) Prove: $P_{n+1}-P_{n}=-\frac{1}{2}\left(P_{n}-P_{n-1}\right)(2 \leq n \leq 99)$;
(3) Find the values of $P_{99}, P_{100}$.
(1) The chess piece can reach station 3 through the following three paths: tossing heads three times in a row, with a probability of $\frac{1}{8}$; tossing tails on the first toss and heads on the second, with a probability of $\frac{1}{4}$; tossing heads on the first toss and tails on the second, with a probability of $\frac{1}{4}$. Therefore, $P_{3}=\frac{5}{8}$.
(2) It is easy to see that the chess piece first jumps to station $n-2$ and then tosses tails, with a probability of $\frac{1}{2} P_{n-2}$; the chess piece first jumps to station $n-1$ and then tosses heads, with a probability of $\frac{1}{2} P_{n-1}$. Therefore, we have $P_{n}=\frac{1}{2}\left(P_{n-1}+P_{n-2}\right)$, which means
$$
P_{n}-P_{n-1}=-\frac{1}{2}\left(P_{n-1}-P_{n-2}\right)
$$

or equivalently,
$$
P_{n+1}-P_{n}=-\frac{1}{2}\left(P_{n}-P_{n-1}\right) \quad(2 \leq n \leq 99)
$$
(3) From (2), we know that the sequence $\left\{P_{n}-P_{n-1}\right\}(n \geq 1)$ is a geometric sequence with the first term $P_{1}-P_{0}=\frac{1}{2}-1=-\frac{1}{2}$ and the common ratio $-\frac{1}{2}$. Therefore, we have
$$
P_{n}-P_{n-1}=\left(-\frac{1}{2}\right)^{n-1}\left(P_{1}-P_{0}\right)=\frac{(-1)^{n}}{2^{n}}
$$

From this, we get
$$
P_{99}=\left(-\frac{1}{2}\right)^{99}+\left(-\frac{1}{2}\right)^{98}+\cdots+\left(-\frac{1}{2}\right)+1=\frac{2}{3}\left(1-\frac{1}{2^{100}}\right)
$$

Since the game automatically stops if the piece reaches station 99, we have $P_{100}=\frac{1}{2} P_{98}=\frac{1}{3}\left(1+\frac{1}{2^{99}}\right)$.
20 points