Points $H$ and $T$ are marked respectively on the sides $BC$ abd $AC$ of triangle $ABC$ so that $AH$ is the altitude and $BT$ is the bisectrix $ABC$. It is known that the gravity center of $ABC$ lies on the line $HT$.
a) Find $AC$ if $BC$=a nad $AB$=c.
b) Determine all possible values of $\frac{c}{b}$ for all triangles $ABC$ satisfying the given condition.
1. **Given Information and Setup:**
   - Points \( H \) and \( T \) are marked on sides \( BC \) and \( AC \) of triangle \( ABC \) respectively.
   - \( AH \) is the altitude from \( A \) to \( BC \).
   - \( BT \) is the angle bisector of \( \angle ABC \).
   - The centroid \( G \) of \( \triangle ABC \) lies on the line \( HT \).
   - We need to find \( AC \) if \( BC = a \) and \( AB = c \).
   - We also need to determine all possible values of \( \frac{c}{b} \) for all triangles \( ABC \) satisfying the given condition.

2. **Using the Second Median Theorem:**
   - The second median theorem states that the length of the median from a vertex to the midpoint of the opposite side can be expressed in terms of the sides of the triangle.
   - For \( \triangle ABC \), let \( M \) be the midpoint of \( BC \). Then, the length of the median \( AM \) is given by:
     \[
     AM = \sqrt{\frac{2b^2 + 2c^2 - a^2}{4}}
     \]

3. **Altitude and Angle Bisector:**
   - The altitude \( AH \) from \( A \) to \( BC \) is perpendicular to \( BC \).
   - The angle bisector \( BT \) divides \( \angle ABC \) into two equal angles.

4. **Using Menelaus' Theorem:**
   - Menelaus' theorem for triangle \( AMC \) with transversal \( MGA \) states:
     \[
     \frac{HM}{HC} \cdot \frac{CT}{TA} \cdot \frac{AG}{GM} = 1
     \]
   - Given that \( G \) is the centroid, \( AG:GM = 2:1 \).

5. **Expressing \( HM \) and \( HC \):**
   - From the second median theorem, we have:
     \[
     HM = \frac{b^2 - c^2}{2a}
     \]
   - The length \( HC \) can be expressed as:
     \[
     HC = \frac{b^2 - c^2}{2a} + \frac{a}{2} = \frac{b^2 + a^2 - c^2}{2a}
     \]

6. **Applying Menelaus' Theorem:**
   - Substituting the expressions for \( HM \) and \( HC \) into Menelaus' theorem:
     \[
     \frac{HM}{HC} \cdot \frac{CT}{TA} \cdot \frac{AG}{GM} = 1
     \]
     \[
     \frac{\frac{b^2 - c^2}{2a}}{\frac{b^2 + a^2 - c^2}{2a}} \cdot \frac{a}{c} \cdot 2 = 1
     \]
     \[
     \frac{b^2 - c^2}{b^2 + a^2 - c^2} \cdot \frac{a}{c} \cdot 2 = 1
     \]

7. **Solving for \( b \):**
   - Simplifying the equation:
     \[
     \frac{(b^2 - c^2) \cdot a \cdot 2}{(b^2 + a^2 - c^2) \cdot c} = 1
     \]
     \[
     \frac{2a(b^2 - c^2)}{c(b^2 + a^2 - c^2)} = 1
     \]
     \[
     2a(b^2 - c^2) = c(b^2 + a^2 - c^2)
     \]
     \[
     2ab^2 - 2ac^2 = cb^2 + ca^2 - cc^2
     \]
     \[
     2ab^2 - cb^2 = ca^2 - cc^2 + 2ac^2
     \]
     \[
     b^2(2a - c) = c(a^2 + 2ac - c^2)
     \]
     \[
     b^2 = \frac{c(a^2 + 2ac - c^2)}{2a - c}
     \]
     \[
     b = \sqrt{\frac{c(a^2 + 2ac - c^2)}{2a - c}}
     \]

The final answer is \( \boxed{ b = \sqrt{\frac{c(a^2 + 2ac - c^2)}{2a - c}} } \)