In Euclidean space, take the point $ N(0,\ 0,\ 1)$ on the sphere $ S$ with radius 1 centered in the origin. For moving points $ P,\ Q$ on $ S$ such that $ NP \equal{} NQ$ and $ \angle{PNQ} \equal{} \theta \left(0 < \theta < \frac {\pi}{2}\right),$ consider the solid figure $ T$ in which the line segment $ PQ$ can be passed.

(1) Show that $ z$ coordinates of $ P,\ Q$ are equal.

(2) When $ P$ is on the palne $ z \equal{} h,$ express the length of $ PQ$ in terms of $ \theta$ and $ h$.

(3) Draw the outline of the cross section by cutting $ T$ by the plane $ z \equal{} h$, then express the area in terms of $ \theta$ and $ h$.

(4) Pay attention to the range for which $ h$ can be valued, express the volume $ V$ of $ T$ in terms of $ \theta$, then  find the maximum $ V$ when let $ \theta$ vary.
### Part (1)
1. Given the point \( N(0, 0, 1) \) on the sphere \( S \) with radius 1 centered at the origin, and points \( P \) and \( Q \) on \( S \) such that \( NP = NQ \) and \( \angle{PNQ} = \theta \).
2. Since \( NP = NQ \), points \( P \) and \( Q \) are equidistant from \( N \). This implies that \( P \) and \( Q \) lie on a circle centered at \( N \) on the sphere.
3. The circle lies in a plane parallel to the \( xy \)-plane because \( N \) is on the \( z \)-axis.
4. Therefore, the \( z \)-coordinates of \( P \) and \( Q \) must be equal.

\[
\boxed{z_P = z_Q}
\]

### Part (2)
1. Let \( P \) be on the plane \( z = h \). Since \( P \) lies on the sphere, its coordinates can be written as \( P(x, y, h) \).
2. The distance from \( N \) to \( P \) is given by:
   \[
   NP = \sqrt{x^2 + y^2 + (h - 1)^2}
   \]
3. Since \( NP = 1 \) (radius of the sphere), we have:
   \[
   \sqrt{x^2 + y^2 + (h - 1)^2} = 1
   \]
   Squaring both sides:
   \[
   x^2 + y^2 + (h - 1)^2 = 1
   \]
4. The length of \( PQ \) can be found using the fact that \( \angle{PNQ} = \theta \). The arc length \( PQ \) on the circle of radius \( r \) is given by:
   \[
   PQ = 2r \sin\left(\frac{\theta}{2}\right)
   \]
5. The radius \( r \) of the circle at height \( h \) is:
   \[
   r = \sqrt{1 - h^2}
   \]
6. Therefore, the length of \( PQ \) is:
   \[
   PQ = 2 \sqrt{1 - h^2} \sin\left(\frac{\theta}{2}\right)
   \]

\[
\boxed{PQ = 2 \sqrt{1 - h^2} \sin\left(\frac{\theta}{2}\right)}
\]

### Part (3)
1. The cross-section of \( T \) by the plane \( z = h \) is a circle with radius \( r = \sqrt{1 - h^2} \).
2. The area of this circle is:
   \[
   A = \pi r^2 = \pi (1 - h^2)
   \]
3. The area of the cross-section is:
   \[
   A = \pi (1 - h^2)
   \]

\[
\boxed{A = \pi (1 - h^2)}
\]

### Part (4)
1. The range for \( h \) is determined by the fact that \( P \) and \( Q \) must lie on the sphere. Thus, \( -1 \leq h \leq 1 \).
2. The volume \( V \) of \( T \) can be found by integrating the area of the cross-sections from \( h = -1 \) to \( h = 1 \):
   \[
   V = \int_{-1}^{1} \pi (1 - h^2) \, dh
   \]
3. Evaluating the integral:
   \[
   V = \pi \int_{-1}^{1} (1 - h^2) \, dh = \pi \left[ h - \frac{h^3}{3} \right]_{-1}^{1}
   \]
4. Substituting the limits:
   \[
   V = \pi \left( \left[ 1 - \frac{1}{3} \right] - \left[ -1 + \frac{1}{3} \right] \right) = \pi \left( \frac{2}{3} + \frac{2}{3} \right) = \frac{4\pi}{3}
   \]
5. The volume \( V \) is independent of \( \theta \) and is maximized when \( \theta = \frac{\pi}{2} \).

\[
\boxed{V = \frac{4\pi}{3}}
\]