Problem 1257

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 $\boxed{\frac{15}{16}V}$.