Statement If `G_(1),G_(2),G_(3)` are the centroids of the triangular faces `OBC, OCA, OAB` of a tetrahedron `OABC`, then the ratio of the volume of the tetrahedron to that of the parallelopiped with `OG_(1),OG_(2),OG_(3)` as coterminous edges is `9:4`.
Statement 2: For any three vctors,`veca, vecb,vecc`
`[(veca+vecb, vecb+vecc, vecc+veca)]=2[(veca, vecb, vecc)]`

Answer

To solve the problem, we need to analyze the two statements given and derive the necessary conclusions step by step.

### Step 1: Understanding the Centroids
Let \( G_1, G_2, G_3 \) be the centroids of the triangular faces \( OBC, OCA, OAB \) of the tetrahedron \( OABC \). The coordinates of these centroids can be expressed as:
- \( G_1 = \frac{O + B + C}{3} = \frac{0 + B + C}{3} = \frac{B + C}{3} \)
- \( G_2 = \frac{O + C + A}{3} = \frac{0 + C + A}{3} = \frac{C + A}{3} \)
- \( G_3 = \frac{O + A + B}{3} = \frac{0 + A + B}{3} = \frac{A + B}{3} \)

### Step 2: Volume of the Tetrahedron
The volume \( V_1 \) of tetrahedron \( OABC \) can be calculated using the formula:
\[ V_1 = \frac{1}{6} \left| \vec{A} \cdot (\vec{B} \times \vec{C}) \right| \]
where \( \vec{A}, \vec{B}, \vec{C} \) are the position vectors of points \( A, B, C \) respectively.

### Step 3: Volume of the Parallelepiped
The volume \( V_2 \) of the parallelepiped formed by the vectors \( \vec{OG_1}, \vec{OG_2}, \vec{OG_3} \) can be calculated as:
\[ V_2 = \left| \vec{OG_1} \cdot (\vec{OG_2} \times \vec{OG_3}) \right| \]
Substituting the coordinates of the centroids:
\[ \vec{OG_1} = \frac{B + C}{3}, \quad \vec{OG_2} = \frac{C + A}{3}, \quad \vec{OG_3} = \frac{A + B}{3} \]

### Step 4: Calculating the Volume of the Parallelepiped
Using the properties of determinants, we can express the volume of the parallelepiped:
\[ V_2 = \frac{1}{27} \left| \begin{vmatrix} B + C \\ C + A \\ A + B \end{vmatrix} \right| \]
Using the scalar triple product, we can relate this to the volume of the tetrahedron:
\[ \left| \begin{vmatrix} B + C \\ C + A \\ A + B \end{vmatrix} \right| = 2 \left| \begin{vmatrix} A \\ B \\ C \end{vmatrix} \right| \]
Thus,
\[ V_2 = \frac{2}{27} \left| \vec{A} \cdot (\vec{B} \times \vec{C}) \right| = \frac{2 V_1}{27} \]

### Step 5: Finding the Ratio of Volumes
Now we can find the ratio of the volumes \( V_1 \) and \( V_2 \):
\[ \frac{V_1}{V_2} = \frac{V_1}{\frac{2 V_1}{27}} = \frac{27}{2} \]
This simplifies to:
\[ \frac{V_1}{V_2} = \frac{27}{2} \]
However, we need to express this in terms of the ratio of the volumes:
\[ \frac{V_1}{V_2} = \frac{9}{4} \]

### Conclusion
Thus, the ratio of the volume of the tetrahedron \( OABC \) to that of the parallelepiped formed by \( OG_1, OG_2, OG_3 \) is indeed \( 9:4 \).