Consider a rectangular parallelepiped $ABCDEFGH$.
(a) If this parallelepiped is a cube, what is the measure of the angle $\angle FCH$ ?
(b) Can the lengths of the edges of this parallelepiped be chosen so that the angle $\angle FCH =45^o$?
(c) What are all the values that the measure of the angle $\angle FCH$ can take ? 
[img]https://1.bp.blogspot.com/-5IVxknY90No/XzAOpAgAfnI/AAAAAAAAMTQ/Zbmx0oFEwz4lWvJHmQalectcN6o4SM3yACLcBGAsYHQ/s0/OBM%2B2012%2Bp1.jpg[/img]
### Part (a)
If the parallelepiped is a cube, all edges are of equal length, say \( s \). The vertices \( F, C, \) and \( H \) form a right triangle in the cube. 

1. **Identify the coordinates of the vertices:**
   - Let \( A = (0, 0, 0) \), \( B = (s, 0, 0) \), \( C = (s, s, 0) \), \( D = (0, s, 0) \)
   - \( E = (0, 0, s) \), \( F = (s, 0, s) \), \( G = (s, s, s) \), \( H = (0, s, s) \)

2. **Calculate the vectors \( \overrightarrow{FC} \) and \( \overrightarrow{FH} \):**
   - \( \overrightarrow{FC} = (s, s, -s) \)
   - \( \overrightarrow{FH} = (-s, s, s) \)

3. **Use the dot product to find the angle \( \angle FCH \):**
   \[
   \overrightarrow{FC} \cdot \overrightarrow{FH} = s(-s) + s(s) + (-s)s = -s^2 + s^2 - s^2 = -s^2
   \]
   \[
   \|\overrightarrow{FC}\| = \|\overrightarrow{FH}\| = \sqrt{s^2 + s^2 + s^2} = s\sqrt{3}
   \]
   \[
   \cos \theta = \frac{\overrightarrow{FC} \cdot \overrightarrow{FH}}{\|\overrightarrow{FC}\| \|\overrightarrow{FH}\|} = \frac{-s^2}{s^3} = -\frac{1}{3}
   \]
   \[
   \theta = \cos^{-1}\left(-\frac{1}{3}\right)
   \]

Thus, the measure of the angle \( \angle FCH \) in a cube is \( \cos^{-1}\left(-\frac{1}{3}\right) \).

### Part (b)
We need to determine if the lengths of the edges of the parallelepiped can be chosen such that \( \angle FCH = 45^\circ \).

1. **Using the Law of Cosines in 3D:**
   \[
   \cos \theta = \frac{\overrightarrow{FC} \cdot \overrightarrow{FH}}{\|\overrightarrow{FC}\| \|\overrightarrow{FH}\|}
   \]
   For \( \theta = 45^\circ \):
   \[
   \cos 45^\circ = \frac{1}{\sqrt{2}}
   \]

2. **Set up the equation:**
   \[
   \frac{\overrightarrow{FC} \cdot \overrightarrow{FH}}{\|\overrightarrow{FC}\| \|\overrightarrow{FH}\|} = \frac{1}{\sqrt{2}}
   \]

3. **Solve for the edge lengths:**
   \[
   \overrightarrow{FC} = (a, b, -c), \quad \overrightarrow{FH} = (-a, b, c)
   \]
   \[
   \overrightarrow{FC} \cdot \overrightarrow{FH} = a(-a) + b(b) + (-c)c = -a^2 + b^2 - c^2
   \]
   \[
   \|\overrightarrow{FC}\| = \sqrt{a^2 + b^2 + c^2}, \quad \|\overrightarrow{FH}\| = \sqrt{a^2 + b^2 + c^2}
   \]
   \[
   \frac{-a^2 + b^2 - c^2}{a^2 + b^2 + c^2} = \frac{1}{\sqrt{2}}
   \]

This equation can be solved for specific values of \( a, b, \) and \( c \). Therefore, it is possible to choose the lengths of the edges such that \( \angle FCH = 45^\circ \).

### Part (c)
To find all possible values of \( \angle FCH \):

1. **Generalize the dot product and magnitudes:**
   \[
   \overrightarrow{FC} = (a, b, -c), \quad \overrightarrow{FH} = (-a, b, c)
   \]
   \[
   \overrightarrow{FC} \cdot \overrightarrow{FH} = -a^2 + b^2 - c^2
   \]
   \[
   \|\overrightarrow{FC}\| = \|\overrightarrow{FH}\| = \sqrt{a^2 + b^2 + c^2}
   \]
   \[
   \cos \theta = \frac{-a^2 + b^2 - c^2}{a^2 + b^2 + c^2}
   \]

2. **Determine the range of \( \cos \theta \):**
   \[
   -1 \leq \frac{-a^2 + b^2 - c^2}{a^2 + b^2 + c^2} \leq 1
   \]

3. **Solve for \( \theta \):**
   \[
   \theta = \cos^{-1}\left(\frac{-a^2 + b^2 - c^2}{a^2 + b^2 + c^2}\right)
   \]

The angle \( \theta \) can take any value in the range \( [0^\circ, 180^\circ] \) depending on the values of \( a, b, \) and \( c \).

The final answer is \( \boxed{ \cos^{-1}\left(-\frac{1}{3}\right) } \) for part (a), \( 45^\circ \) for part (b), and any value in \( [0^\circ, 180^\circ] \) for part (c).