Let $M$ be an interior point of tetrahedron $V ABC$. Denote by $A_1,B_1, C_1$ the points of intersection of lines $MA,MB,MC$ with the planes $VBC,V CA,V AB$, and by $A_2,B_2, C_2$ the points of intersection of lines $V A_1, VB_1, V C_1$ with the sides $BC,CA,AB$.

[b](a)[/b] Prove that the volume of the tetrahedron $V A_2B_2C_2$ does not exceed one-fourth of the volume of $V ABC$.

[b](b)[/b] Calculate the volume of the tetrahedron $V_1A_1B_1C_1$ as a function of the volume of $V ABC$, where $V_1$ is the point of intersection of the line $VM$ with the plane $ABC$, and $M$ is the barycenter of $V ABC$.
**(a)** Prove that the volume of the tetrahedron \( V A_2B_2C_2 \) does not exceed one-fourth of the volume of \( V ABC \).

1. **Identify Midpoints**: Let \( A_0, B_0, C_0 \) be the midpoints of the edges \( BC, CA, AB \) respectively. 

2. **Area Comparison**: The area of triangle \( A_2B_2C_2 \) is less than or equal to the area of triangle \( A_0B_0C_0 \). This is because \( A_0, B_0, C_0 \) are midpoints, and the triangle formed by midpoints of a triangle has an area that is one-fourth of the original triangle's area:
   \[
   S_{\triangle A_2B_2C_2} \leq S_{\triangle A_0B_0C_0} = \frac{1}{4} S_{\triangle ABC}
   \]

3. **Height Comparison**: The height from \( V \) to the plane \( ABC \) is the same as the height from \( V \) to the plane \( A_2B_2C_2 \):
   \[
   d(V, (ABC)) = d(V, (A_2B_2C_2))
   \]

4. **Volume Calculation**: The volume of a tetrahedron is given by:
   \[
   V_{VA_2B_2C_2} = \frac{1}{3} \times \text{height} \times \text{base area}
   \]
   Using the height and area relationships:
   \[
   V_{VA_2B_2C_2} = \frac{1}{3} \times d(V, (A_2B_2C_2)) \times S_{\triangle A_2B_2C_2} \leq \frac{1}{3} \times d(V, (ABC)) \times \frac{1}{4} S_{\triangle ABC}
   \]
   Simplifying, we get:
   \[
   V_{VA_2B_2C_2} \leq \frac{1}{4} V_{VABC}
   \]

Thus, the volume of the tetrahedron \( V A_2B_2C_2 \) does not exceed one-fourth of the volume of \( V ABC \).

\(\blacksquare\)

**(b)** Calculate the volume of the tetrahedron \( V_1A_1B_1C_1 \) as a function of the volume of \( V ABC \), where \( V_1 \) is the point of intersection of the line \( VM \) with the plane \( ABC \), and \( M \) is the barycenter of \( V ABC \).

1. **Barycenter Property**: The barycenter \( M \) of a tetrahedron divides each median in the ratio \( 3:1 \). Therefore, the point \( V_1 \) is such that \( VM = 3 \times MV_1 \).

2. **Volume Ratio**: The volume of a tetrahedron formed by a point inside the tetrahedron and the faces of the original tetrahedron is proportional to the ratio of the segments formed by the point. Since \( M \) is the barycenter, the volume of \( V_1A_1B_1C_1 \) is:
   \[
   V_{V_1A_1B_1C_1} = \frac{1}{4} V_{VABC}
   \]

Thus, the volume of the tetrahedron \( V_1A_1B_1C_1 \) is one-fourth of the volume of \( V ABC \).

\(\blacksquare\)

The final answer is \( \boxed{ \frac{1}{4} V_{VABC} } \)