In the cuboid $ABCDA'B'C'D'$ with $AB=a$, $AD=b$ and $AA'=c$ such that $a>b>c>0$, the points $E$ and $F$ are the orthogonal projections of $A$ on the lines $A'D$ and $A'B$, respectively, and the points $M$ and $N$ are the orthogonal projections of $C$ on the lines $C'D$ and $C'B$, respectively. Let $DF\cap BE=\{G\}$ and $DN\cap BM=\{P\}$.
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[*] Show that $(A'AG)\parallel (C'CP)$ and determine the distance between these two planes;
[*] Show that $GP\parallel (ABC)$ and determine the distance between the line $GP$ and the plane $(ABC)$.
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[i]Petre Simion, Nicolae Victor Ioan[/i]
### Part (a)

1. **Identify the orthogonal projections:**
   - \( E \) is the orthogonal projection of \( A \) on \( A'D \).
   - \( F \) is the orthogonal projection of \( A \) on \( A'B \).
   - \( M \) is the orthogonal projection of \( C \) on \( C'D \).
   - \( N \) is the orthogonal projection of \( C \) on \( C'B \).

2. **Determine the orthocenters:**
   - Since \( (ABE) \perp A'D \), it follows that \( BE \perp A'D \). Similarly, \( G \) is the orthocenter of \( \triangle A'DB \).
   - Similarly, \( P \) is the orthocenter of \( \triangle BC'D \).

3. **Use the reverse of the three perpendiculars theorem:**
   - From the orthocenter properties, \( AG \perp (A'DB) \) and \( CP \perp (BC'D) \).

4. **Parallel planes:**
   - Since \( BD \perp (A'AG) \) and \( BD \perp (C'CP) \), it implies that \( (A'AG) \parallel (C'CP) \).

5. **Calculate the distance between the planes:**
   - Let \( X \in A'G \cap BD \) and \( Y \in C'P \cap BD \).
   - Using the Pythagorean theorem, the distance \( BD = \sqrt{a^2 + b^2} \).
   - The distance between the planes is given by:
     \[
     \mathrm{dist}((A'AG), (C'CP)) = XY = BD - 2BY = \sqrt{a^2 + b^2} - \frac{2b^2}{\sqrt{a^2 + b^2}} = \frac{a^2 - b^2}{\sqrt{a^2 + b^2}}.
     \]

### Part (b)

1. **Parallelism of \( GP \) and \( (ABC) \):**
   - Let \( (ABC) = \alpha \), \( Q = \mathrm{pr}_\alpha(G) \), and \( R = \mathrm{pr}_\alpha(P) \).
   - From \( \triangle A'BD \cong \triangle C'DB \), it follows that \( \frac{XG}{XA'} = \frac{YP}{YC'} \).
   - Using similar triangles \( \triangle XGQ \sim \triangle XAA' \) and \( \triangle YPR \sim \triangle YC'C \), we get:
     \[
     \frac{XG}{XA'} = \frac{GQ}{c} \quad \text{and} \quad \frac{YP}{YC'} = \frac{PR}{c}.
     \]
   - Therefore, \( GQ = PR \), which implies \( GP \parallel \alpha \).

2. **Calculate the distance \( PR \):**
   - Using the formulas for the leg and height of a right-angled triangle:
     \[
     YC = \frac{ab}{\sqrt{a^2 + b^2}}, \quad YC' = \frac{\sqrt{a^2b^2 + b^2c^2 + c^2a^2}}{\sqrt{a^2 + b^2}}, \quad CP = \frac{abc}{\sqrt{a^2b^2 + b^2c^2 + c^2a^2}},
     \]
     \[
     YP = \frac{a^2b^2}{\sqrt{a^2 + b^2} \cdot \sqrt{a^2b^2 + b^2c^2 + c^2a^2}}.
     \]
   - Thus,
     \[
     PR = \frac{PC \cdot PY}{YC} = \frac{a^2b^2c}{a^2b^2 + b^2c^2 + c^2a^2}.
     \]

The final answer is \( \boxed{ \frac{a^2 - b^2}{\sqrt{a^2 + b^2}} } \) for part (a) and \(\frac{a^2b^2c}{a^2b^2 + b^2c^2 + c^2a^2}\) for part (b).