Let $ H$ be the piont of midpoint of the cord $ PQ$ that is on the circle centered the origin $ O$ with radius $ 1.$
Suppose the length of the cord $ PQ$ is $ 2\sin \frac {t}{2}$ for the angle $ t\ (0\leq t\leq \pi)$ that is formed by half-ray $ OH$ and the positive direction of the $ x$ axis. Answer the following questions.

(1) Express the coordiante of $ H$ in terms of $ t$.

(2) When $ t$ moves in the range of $ 0\leq t\leq \pi$, find the minimum value of $ x$ coordinate of $ H$.

(3) When $ t$ moves in the range of $ 0\leq t\leq \frac {\pi}{2}$, find the area $ S$ of the region bounded by the curve drawn by the point $ H$ and the $ x$ axis and the $ y$ axis.
### Part (1)
To find the coordinates of \( H \) in terms of \( t \):

1. Given the length of the chord \( PQ \) is \( 2\sin\frac{t}{2} \), the distance from \( H \) to \( P \) or \( Q \) is half of this length:
   \[
   PH = \sin\frac{t}{2}
   \]

2. Since \( H \) is the midpoint of the chord \( PQ \), it lies on the circle centered at the origin \( O \) with radius 1. The distance \( OH \) can be found using the Pythagorean theorem:
   \[
   OH = \sqrt{OP^2 - PH^2} = \sqrt{1 - \sin^2\frac{t}{2}} = \cos\frac{t}{2}
   \]

3. The coordinates of \( H \) can be expressed in terms of \( t \) as:
   \[
   H = \left( \cos\frac{t}{2} \cos t, \cos\frac{t}{2} \sin t \right)
   \]

Thus, the coordinates of \( H \) are:
\[
\boxed{H \equiv \left( \cos\frac{t}{2} \cos t, \cos\frac{t}{2} \sin t \right)}
\]

### Part (2)
To find the minimum value of the \( x \)-coordinate of \( H \) when \( t \) ranges from \( 0 \) to \( \pi \):

1. The \( x \)-coordinate of \( H \) is:
   \[
   x = \cos\frac{t}{2} \cos t
   \]

2. To find the critical points, we take the derivative of \( x \) with respect to \( t \):
   \[
   \frac{dx}{dt} = -\frac{1}{2} \sin\frac{t}{2} \cos t - \cos\frac{t}{2} \sin t
   \]

3. Simplify the derivative:
   \[
   \frac{dx}{dt} = -\frac{1}{2} \sin\frac{t}{2} \cos t - \cos\frac{t}{2} \sin t = -\frac{1}{2} \sin\frac{t}{2} (1 - 2\sin^2\frac{t}{2}) - 2\sin\frac{t}{2} \cos^2\frac{t}{2}
   \]

4. Factor out \( \sin\frac{t}{2} \):
   \[
   \frac{dx}{dt} = \sin\frac{t}{2} \left( 3\sin^2\frac{t}{2} - \frac{5}{2} \right)
   \]

5. Set the derivative to zero to find critical points:
   \[
   3\sin^2\frac{t}{2} - \frac{5}{2} = 0 \implies \sin^2\frac{t}{2} = \frac{5}{6}
   \]

6. Solve for \( t \):
   \[
   \sin\frac{t}{2} = \sqrt{\frac{5}{6}} \implies t = 2\arcsin\sqrt{\frac{5}{6}}
   \]

7. Substitute \( t \) back into the \( x \)-coordinate expression to find the minimum value:
   \[
   x = \cos\frac{t}{2} \cos t = \cos\left( \arcsin\sqrt{\frac{5}{6}} \right) \cos\left( 2\arcsin\sqrt{\frac{5}{6}} \right)
   \]

   Using trigonometric identities, we find:
   \[
   \cos\left( \arcsin\sqrt{\frac{5}{6}} \right) = \sqrt{1 - \left( \sqrt{\frac{5}{6}} \right)^2} = \sqrt{\frac{1}{6}}
   \]

   And:
   \[
   \cos\left( 2\arcsin\sqrt{\frac{5}{6}} \right) = 1 - 2\left( \sqrt{\frac{5}{6}} \right)^2 = 1 - \frac{10}{6} = -\frac{1}{3}
   \]

   Therefore:
   \[
   x = \sqrt{\frac{1}{6}} \cdot \left( -\frac{1}{3} \right) = -\frac{1}{3\sqrt{6}}
   \]

Thus, the minimum \( x \)-coordinate is:
\[
\boxed{\min x = -\frac{1}{3\sqrt{6}}}
\]

### Part (3)
To find the area \( S \) of the region bounded by the curve drawn by the point \( H \) and the \( x \)-axis and the \( y \)-axis when \( t \) ranges from \( 0 \) to \( \frac{\pi}{2} \):

1. The area \( S \) can be found using the integral:
   \[
   S = \frac{1}{2} \int_0^{\frac{\pi}{2}} \left( x(t) y'(t) - y(t) x'(t) \right) dt
   \]

2. Substitute \( x(t) = \cos\frac{t}{2} \cos t \) and \( y(t) = \cos\frac{t}{2} \sin t \):
   \[
   S = \frac{1}{2} \int_0^{\frac{\pi}{2}} \left[ \cos\frac{t}{2} \cos t \left( -\frac{1}{2} \sin\frac{t}{2} \sin t + \cos\frac{t}{2} \cos t \right) + \cos\frac{t}{2} \sin t \left( \frac{1}{2} \sin\frac{t}{2} \cos t + \cos\frac{t}{2} \sin t \right) \right] dt
   \]

3. Simplify the integrand:
   \[
   S = \frac{1}{2} \int_0^{\frac{\pi}{2}} \left[ \cos^2\frac{t}{2} \cos^2 t + \cos^2\frac{t}{2} \sin^2 t \right] dt
   \]

4. Factor out \( \cos^2\frac{t}{2} \):
   \[
   S = \frac{1}{2} \int_0^{\frac{\pi}{2}} \cos^2\frac{t}{2} \left( \cos^2 t + \sin^2 t \right) dt = \frac{1}{2} \int_0^{\frac{\pi}{2}} \cos^2\frac{t}{2} \cdot 1 \, dt
   \]

5. Use the identity \( \cos^2\frac{t}{2} = \frac{1 + \cos t}{2} \):
   \[
   S = \frac{1}{2} \int_0^{\frac{\pi}{2}} \frac{1 + \cos t}{2} \, dt = \frac{1}{4} \int_0^{\frac{\pi}{2}} (1 + \cos t) \, dt
   \]

6. Integrate:
   \[
   S = \frac{1}{4} \left[ t + \sin t \right]_0^{\frac{\pi}{2}} = \frac{1}{4} \left( \frac{\pi}{2} + 1 \right)
   \]

Thus, the area \( S \) is:
\[
\boxed{S = \frac{\pi + 2}{8} \text{ square units}}
\]