A regular tetrahedron $SABC$ of volume $V$ is given. The midpoints $D$ and $E$ are taken on $SA$ and $SB$ respectively and the point $F$ is taken on the edge $SC$ such that $SF: FC = 1: 3$. Find the volume of the pentahedron $FDEABC$.
1. **Volume of the Tetrahedron:**
   The volume \( V \) of a regular tetrahedron can be expressed as:
   \[
   V = \frac{1}{3} B \cdot h
   \]
   where \( B \) is the area of the base triangle \( \triangle ABC \) and \( h \) is the height from the apex \( S \) to the base \( \triangle ABC \).

2. **Midpoints and Proportions:**
   - \( D \) is the midpoint of \( SA \), so \( SD = DA = \frac{1}{2}SA \).
   - \( E \) is the midpoint of \( SB \), so \( SE = EB = \frac{1}{2}SB \).
   - \( F \) is on \( SC \) such that \( SF : FC = 1 : 3 \). This means \( SF = \frac{1}{4}SC \) and \( FC = \frac{3}{4}SC \).

3. **Area of \( \triangle DEF' \):**
   - \( F' \) is the midpoint of \( SC \), so \( SF' = F'C = \frac{1}{2}SC \).
   - The area of \( \triangle DEF' \) is \( \frac{1}{4} \) of the area of \( \triangle ABC \) because \( D, E, F' \) are midpoints of the sides of \( \triangle SABC \).
   \[
   \text{Area of } \triangle DEF' = \frac{1}{4} B
   \]

4. **Volume of the Tetrahedron \( SDEF' \):**
   - The height of \( S \) from \( \triangle DEF' \) is \( \frac{h}{2} \) because \( D, E, F' \) are midpoints.
   - The volume of \( SDEF' \) is:
   \[
   V_{SDEF'} = \frac{1}{3} \left( \frac{1}{4} B \right) \left( \frac{h}{2} \right) = \frac{1}{3} \cdot \frac{1}{4} B \cdot \frac{h}{2} = \frac{1}{24} B \cdot h
   \]

5. **Volume of the Tetrahedron \( SFDEF' \):**
   - \( F \) is on \( SC \) such that \( SF = \frac{1}{4}SC \).
   - The volume of \( SFDEF' \) is:
   \[
   V_{SFDEF'} = \frac{1}{2} \cdot \frac{1}{4} B \cdot \frac{h}{4} = \frac{1}{2} \cdot \frac{1}{4} B \cdot \frac{h}{4} = \frac{1}{32} B \cdot h
   \]

6. **Sum of the Volumes:**
   - The total volume of the pentahedron \( FDEABC \) is the volume of the tetrahedron \( SABC \) minus the volume of the tetrahedron \( SFDEF' \):
   \[
   V_{FDEABC} = V - V_{SFDEF'} = V - \frac{1}{32} B \cdot h
   \]
   - Since \( V = \frac{1}{3} B \cdot h \):
   \[
   V_{FDEABC} = \frac{1}{3} B \cdot h - \frac{1}{32} B \cdot h = \frac{32}{96} B \cdot h - \frac{3}{96} B \cdot h = \frac{29}{96} B \cdot h
   \]

The final answer is \(\boxed{\frac{15}{16}V}\).