An unfair coin, which has the probability of $a\ \left(0<a<\frac 12\right)$ for showing $Heads$ and $1-a$ for showing $Tails$, is flipped $n\geq 2$ times. After $n$-th trial, denote by $A_n$ the event that heads are showing on at least two times and by$B_n$ the event that are not showing in the order of $tails\rightarrow heads$, until the trials $T_1,\ T_2,\ \cdots ,\ T_n$ will be finished . Answer the following questions:

(1) Find the probabilities $P(A_n),\ P(B_n)$.

(2) Find the probability $P(A_n\cap B_n )$.

(3) Find the limit $\lim_{n\to\infty} \frac{P(A_n) P(B_n)}{P(A_n\cap B_n )}.$

You may use $\lim_{n\to\infty} nr^n=0\ (0<r<1).$
Let's solve the problem step-by-step.

### Part (1): Find the probabilities \( P(A_n) \) and \( P(B_n) \).

1. **Finding \( P(A_n) \):**

   \( A_n \) is the event that heads show up at least two times in \( n \) trials. 

   - The probability of getting exactly \( k \) heads in \( n \) trials is given by the binomial distribution:
     \[
     P(\text{exactly } k \text{ heads}) = \binom{n}{k} a^k (1-a)^{n-k}
     \]
   - Therefore, the probability of getting at least 2 heads is:
     \[
     P(A_n) = 1 - P(\text{0 heads}) - P(\text{1 head})
     \]
   - Calculating \( P(\text{0 heads}) \) and \( P(\text{1 head}) \):
     \[
     P(\text{0 heads}) = (1-a)^n
     \]
     \[
     P(\text{1 head}) = \binom{n}{1} a (1-a)^{n-1} = n a (1-a)^{n-1}
     \]
   - Thus,
     \[
     P(A_n) = 1 - (1-a)^n - n a (1-a)^{n-1}
     \]

2. **Finding \( P(B_n) \):**

   \( B_n \) is the event that heads do not appear immediately after tails in \( n \) trials.

   - We need to count the sequences where heads do not follow tails. This can be done by considering the sequences where heads appear in blocks, and tails appear in blocks.
   - The probability of such sequences is given by:
     \[
     P(B_n) = \sum_{k=0}^{n} a^k (1-a)^{n-k} \binom{n}{k}
     \]
   - This sum can be simplified using the binomial theorem:
     \[
     P(B_n) = (a + (1-a))^n = 1
     \]
   - However, this is not correct for the specific condition given. Instead, we need to use the generating function approach:
     \[
     P(B_n) = \frac{a^{n+1} - (1-a)^{n+1}}{2a-1}
     \]

### Part (2): Find the probability \( P(A_n \cap B_n) \).

1. **Finding \( P(A_n \cap B_n) \):**

   - We need to find the intersection of the events \( A_n \) and \( B_n \).
   - Using the inclusion-exclusion principle and the given conditions:
     \[
     P(A_n \cap B_n) = P(B_n) - P(\text{0 heads}) - P(\text{1 head})
     \]
   - Thus,
     \[
     P(A_n \cap B_n) = \frac{a^{n+1} - (1-a)^{n+1}}{2a-1} - (1-a)^n - a (1-a)^{n-1}
     \]

### Part (3): Find the limit \( \lim_{n \to \infty} \frac{P(A_n) P(B_n)}{P(A_n \cap B_n)} \).

1. **Finding the limit:**

   - We need to evaluate:
     \[
     \lim_{n \to \infty} \frac{P(A_n) P(B_n)}{P(A_n \cap B_n)}
     \]
   - Substituting the expressions for \( P(A_n) \), \( P(B_n) \), and \( P(A_n \cap B_n) \):
     \[
     \lim_{n \to \infty} \frac{\left(1 - (1-a)^n - n a (1-a)^{n-1}\right) \left(\frac{a^{n+1} - (1-a)^{n+1}}{2a-1}\right)}{\frac{a^{n+1} - (1-a)^{n+1}}{2a-1} - (1-a)^n - a (1-a)^{n-1}}
     \]
   - Simplifying the expression:
     \[
     \lim_{n \to \infty} \frac{\frac{1}{2a-1}}{\frac{1}{2a-1} + \frac{\frac{1}{(1-a)^2}}{1 + \left(\frac{a}{1-a}\right)^{n+1}}}
     \]
   - As \( n \to \infty \), the term \( \left(\frac{a}{1-a}\right)^{n+1} \to 0 \):
     \[
     \lim_{n \to \infty} \frac{1}{1 + \frac{2a-1}{(a-1)^2}} = \left(\frac{1-a}{a}\right)^2
     \]

The final answer is \( \boxed{ \left(\frac{1-a}{a}\right)^2 } \)