[$XYZ$] denotes the area of $\triangle XYZ$
We have a $\triangle ABC$,$BC=6,CA=7,AB=8$
(1)If $O$ is the circumcenter of $\triangle ABC$, find [$OBC$]:[$OCA$]:[$OAB$]
(2)If $H$ is the orthocenter of $\triangle ABC$, find [$HBC$]:[$HCA$]:[$HAB$]
Given a triangle \( \triangle ABC \) with sides \( BC = a = 6 \), \( CA = b = 7 \), and \( AB = c = 8 \), we need to find the ratios of the areas of the triangles formed with the circumcenter \( O \) and the orthocenter \( H \).

### Part 1: Finding the ratio \([OBC] : [OCA] : [OAB]\)

1. **Circumcenter Coordinates**:
   Using barycentric coordinates with respect to \( \triangle ABC \), the circumcenter \( O \) has coordinates:
   \[
   O = (\tan 2A : \tan 2B : \tan 2C)
   \]
   where \( \tan 2A = \frac{2 \tan A}{1 - \tan^2 A} \), and similarly for \( \tan 2B \) and \( \tan 2C \).

2. **Area Ratios**:
   The area of a triangle formed by the circumcenter and two vertices of the original triangle can be expressed in terms of the barycentric coordinates. The ratio of the areas of \( \triangle OBC \), \( \triangle OCA \), and \( \triangle OAB \) is given by the absolute values of the coordinates of \( O \):
   \[
   [OBC] : [OCA] : [OAB] = \left| \tan 2A \right| : \left| \tan 2B \right| : \left| \tan 2C \right|
   \]

3. **Calculating Tangents**:
   Using the Law of Cosines, we find the angles \( A \), \( B \), and \( C \):
   \[
   \cos A = \frac{b^2 + c^2 - a^2}{2bc} = \frac{7^2 + 8^2 - 6^2}{2 \cdot 7 \cdot 8} = \frac{49 + 64 - 36}{112} = \frac{77}{112} = \frac{11}{16}
   \]
   \[
   \cos B = \frac{a^2 + c^2 - b^2}{2ac} = \frac{6^2 + 8^2 - 7^2}{2 \cdot 6 \cdot 8} = \frac{36 + 64 - 49}{96} = \frac{51}{96} = \frac{17}{32}
   \]
   \[
   \cos C = \frac{a^2 + b^2 - c^2}{2ab} = \frac{6^2 + 7^2 - 8^2}{2 \cdot 6 \cdot 7} = \frac{36 + 49 - 64}{84} = \frac{21}{84} = \frac{1}{4}
   \]

   Using these cosines, we can find the tangents:
   \[
   \tan A = \sqrt{\frac{1 - \cos^2 A}{\cos^2 A}} = \sqrt{\frac{1 - \left(\frac{11}{16}\right)^2}{\left(\frac{11}{16}\right)^2}} = \sqrt{\frac{1 - \frac{121}{256}}{\frac{121}{256}}} = \sqrt{\frac{135}{121}} = \frac{\sqrt{135}}{11}
   \]
   Similarly, we can find \( \tan B \) and \( \tan C \).

4. **Final Ratios**:
   The ratios of the areas are:
   \[
   [OBC] : [OCA] : [OAB] = \left| \tan 2A \right| : \left| \tan 2B \right| : \left| \tan 2C \right|
   \]

### Part 2: Finding the ratio \([HBC] : [HCA] : [HAB]\)

1. **Orthocenter Coordinates**:
   Using barycentric coordinates with respect to \( \triangle ABC \), the orthocenter \( H \) has coordinates:
   \[
   H = (\tan A : \tan B : \tan C)
   \]

2. **Area Ratios**:
   The area of a triangle formed by the orthocenter and two vertices of the original triangle can be expressed in terms of the barycentric coordinates. The ratio of the areas of \( \triangle HBC \), \( \triangle HCA \), and \( \triangle HAB \) is given by the absolute values of the coordinates of \( H \):
   \[
   [HBC] : [HCA] : [HAB] = \left| \tan A \right| : \left| \tan B \right| : \left| \tan C \right|
   \]

3. **Final Ratios**:
   The ratios of the areas are:
   \[
   [HBC] : [HCA] : [HAB] = \left| \tan A \right| : \left| \tan B \right| : \left| \tan C \right|
   \]

The final answer is \( \boxed{ [OBC] : [OCA] : [OAB] = \left| \tan 2A \right| : \left| \tan 2B \right| : \left| \tan 2C \right| } \) and \([HBC] : [HCA] : [HAB] = \left| \tan A \right| : \left| \tan B \right| : \left| \tan C \right|\)