Let $l$ be the tangent line at the point $P(s,\ t)$ on a circle $C:x^2+y^2=1$. Denote by $m$ the line passing through the point $(1,\ 0)$, parallel to $l$. Let the line $m$ intersects the circle $C$ at $P'$ other than the point $(1,\ 0)$.

Note : if $m$ is the line $x=1$, then $P'$ is considered as $(1,\ 0)$.

Call $T$ the operation such that the point $P'(s',\ t')$ is obtained from the point $P(s,\ t)$ on $C$. 

(1) Express $s',\ t'$ as the polynomials of $s$ and $t$ respectively.

(2) Let $P_n$ be the point obtained by $n$ operations of $T$ for $P$.

For $P\left(\frac{\sqrt{3}}{2},\ \frac{1}{2}\right)$, plot the points $P_1,\ P_2$ and $P_3$.

(3) For a positive integer $n$, find the number of $P$ such that $P_n=P$.
Given the circle \( C: x^2 + y^2 = 1 \) and a point \( P(s, t) \) on the circle, we need to find the tangent line at \( P \) and the line \( m \) passing through \( (1, 0) \) parallel to the tangent line. We then find the intersection of \( m \) with the circle at a point \( P' \) other than \( (1, 0) \).

1. **Express \( s' \) and \( t' \) as polynomials of \( s \) and \( t \):**

   The tangent line at \( P(s, t) \) on the circle \( x^2 + y^2 = 1 \) has the equation:
   \[
   sx + ty = 1
   \]
   The line \( m \) passing through \( (1, 0) \) and parallel to the tangent line has the same slope, so its equation is:
   \[
   sx + ty = s
   \]
   To find the intersection of this line with the circle \( x^2 + y^2 = 1 \), substitute \( y = \frac{s - sx}{t} \) into the circle's equation:
   \[
   x^2 + \left( \frac{s - sx}{t} \right)^2 = 1
   \]
   Simplifying, we get:
   \[
   x^2 + \frac{s^2 - 2s^2x + s^2x^2}{t^2} = 1
   \]
   \[
   x^2 + \frac{s^2}{t^2} - \frac{2s^2x}{t^2} + \frac{s^2x^2}{t^2} = 1
   \]
   \[
   t^2x^2 + s^2 - 2s^2x + s^2x^2 = t^2
   \]
   \[
   (t^2 + s^2)x^2 - 2s^2x + s^2 - t^2 = 0
   \]
   This is a quadratic equation in \( x \). Solving for \( x \) using the quadratic formula \( x = \frac{-b \pm \sqrt{b^2 - 4ac}}{2a} \):
   \[
   x = \frac{2s^2 \pm \sqrt{4s^4 - 4(t^2 + s^2)(s^2 - t^2)}}{2(t^2 + s^2)}
   \]
   Simplifying inside the square root:
   \[
   x = \frac{2s^2 \pm \sqrt{4s^4 - 4(s^4 - t^4)}}{2(t^2 + s^2)}
   \]
   \[
   x = \frac{2s^2 \pm \sqrt{4t^4}}{2(t^2 + s^2)}
   \]
   \[
   x = \frac{2s^2 \pm 2t^2}{2(t^2 + s^2)}
   \]
   \[
   x = \frac{s^2 \pm t^2}{t^2 + s^2}
   \]
   Since \( x = 1 \) is one solution, the other solution is:
   \[
   x = \frac{s^2 - t^2}{t^2 + s^2}
   \]
   For \( y \):
   \[
   y = \frac{s - sx}{t}
   \]
   Substituting \( x = \frac{s^2 - t^2}{t^2 + s^2} \):
   \[
   y = \frac{s - s \left( \frac{s^2 - t^2}{t^2 + s^2} \right)}{t}
   \]
   \[
   y = \frac{s(t^2 + s^2) - s(s^2 - t^2)}{t(t^2 + s^2)}
   \]
   \[
   y = \frac{st^2 + s^3 - s^3 + st^2}{t(t^2 + s^2)}
   \]
   \[
   y = \frac{2st^2}{t(t^2 + s^2)}
   \]
   \[
   y = \frac{2st^2}{t(t^2 + s^2)}
   \]
   \[
   y = \frac{2st}{t^2 + s^2}
   \]

   Therefore, the coordinates of \( P' \) are:
   \[
   s' = \frac{s^2 - t^2}{t^2 + s^2}, \quad t' = \frac{2st}{t^2 + s^2}
   \]

2. **Plot the points \( P_1, P_2, \) and \( P_3 \) for \( P\left(\frac{\sqrt{3}}{2}, \frac{1}{2}\right) \):**

   Using the formulas derived:
   \[
   P_1\left( \frac{\left(\frac{\sqrt{3}}{2}\right)^2 - \left(\frac{1}{2}\right)^2}{\left(\frac{1}{2}\right)^2 + \left(\frac{\sqrt{3}}{2}\right)^2}, \frac{2 \cdot \frac{\sqrt{3}}{2} \cdot \frac{1}{2}}{\left(\frac{1}{2}\right)^2 + \left(\frac{\sqrt{3}}{2}\right)^2} \right)
   \]
   \[
   P_1\left( \frac{\frac{3}{4} - \frac{1}{4}}{\frac{1}{4} + \frac{3}{4}}, \frac{\sqrt{3}}{2} \cdot \frac{1}{2} \cdot \frac{2}{1} \right)
   \]
   \[
   P_1\left( \frac{1}{2}, \frac{\sqrt{3}}{2} \right)
   \]

   For \( P_2 \):
   \[
   P_2\left( \frac{\left(\frac{1}{2}\right)^2 - \left(\frac{\sqrt{3}}{2}\right)^2}{\left(\frac{\sqrt{3}}{2}\right)^2 + \left(\frac{1}{2}\right)^2}, \frac{2 \cdot \frac{1}{2} \cdot \frac{\sqrt{3}}{2}}{\left(\frac{\sqrt{3}}{2}\right)^2 + \left(\frac{1}{2}\right)^2} \right)
   \]
   \[
   P_2\left( \frac{\frac{1}{4} - \frac{3}{4}}{\frac{3}{4} + \frac{1}{4}}, \frac{\sqrt{3}}{2} \cdot \frac{1}{2} \cdot \frac{2}{1} \right)
   \]
   \[
   P_2\left( -\frac{1}{2}, \frac{\sqrt{3}}{2} \right)
   \]

   For \( P_3 \):
   \[
   P_3\left( \frac{\left(-\frac{1}{2}\right)^2 - \left(\frac{\sqrt{3}}{2}\right)^2}{\left(\frac{\sqrt{3}}{2}\right)^2 + \left(-\frac{1}{2}\right)^2}, \frac{2 \cdot -\frac{1}{2} \cdot \frac{\sqrt{3}}{2}}{\left(\frac{\sqrt{3}}{2}\right)^2 + \left(-\frac{1}{2}\right)^2} \right)
   \]
   \[
   P_3\left( \frac{\frac{1}{4} - \frac{3}{4}}{\frac{3}{4} + \frac{1}{4}}, \frac{\sqrt{3}}{2} \cdot -\frac{1}{2} \cdot \frac{2}{1} \right)
   \]
   \[
   P_3\left( -\frac{1}{2}, \frac{\sqrt{3}}{2} \right)
   \]

   Therefore, the points are:
   \[
   P_1\left( \frac{1}{2}, \frac{\sqrt{3}}{2} \right), \quad P_2\left( -\frac{1}{2}, \frac{\sqrt{3}}{2} \right), \quad P_3\left( -\frac{1}{2}, \frac{\sqrt{3}}{2} \right)
   \]

3. **Find the number of \( P \) such that \( P_n = P \):**

   Considering the complex number representation, let \( z = e^{i\theta} \) where \( \theta \) is the argument of \( P \). The operation \( T \) corresponds to squaring the complex number, so after \( n \) operations, we have:
   \[
   z^{2^n} = z
   \]
   This implies:
   \[
   z^{2^n - 1} = 1
   \]
   The solutions to this equation are the \( 2^n - 1 \) roots of unity, excluding \( z = 1 \). Therefore, there are \( 2^n - 1 \) such points.

The final answer is \( \boxed{ 2^n - 1 } \)