`{:(Column -I,Column -II),((A)"Anti logarithm of" (0.bar(6))"to the base 27 has the value equal to",(P)5),((B)"Characteristic of the logarithm of 2008 to the base 2 is",(Q)7),((C)"The value of b satisfying the equation",(R)9),(log_(e)2.log_(b)625=log_(10)16.log_(e)10 is,),((D)"Number of naughts after decimal before a significant figure",(S)10),("comes in the number" ((5)/(6))^(100)is,):}`

Answer

To solve the given problem, we need to analyze the equations and find the required values step by step.

### Step 1: Understand the equation
We are given the equation:
\[ 2a^2 + 3b^2 = 35 \]
where \( a \) and \( b \) are integers (denoted by \( Z \)).

### Step 2: Rearranging the equation
We can rearrange the equation to express it in a more manageable form:
\[ 3b^2 = 35 - 2a^2 \]
This implies that \( 35 - 2a^2 \) must be non-negative since \( b^2 \) cannot be negative. Therefore:
\[ 35 - 2a^2 \geq 0 \]
This leads to:
\[ 2a^2 \leq 35 \]
or
\[ a^2 \leq 17.5 \]
Thus, \( a \) can take integer values from \( -4 \) to \( 4 \) (since \( 4^2 = 16 \) and \( 5^2 = 25 \) which exceeds 17.5).

### Step 3: Finding possible values for \( a \)
The possible integer values for \( a \) are:
\[ a = -4, -3, -2, -1, 0, 1, 2, 3, 4 \]
This gives us a total of 9 possible values for \( a \).

### Step 4: Finding corresponding values for \( b \)
Now, we will substitute each value of \( a \) into the equation \( 2a^2 + 3b^2 = 35 \) and solve for \( b \).

1. **For \( a = -4 \) or \( a = 4 \)**:
\[ 2(4^2) + 3b^2 = 35 \implies 32 + 3b^2 = 35 \implies 3b^2 = 3 \implies b^2 = 1 \implies b = \pm 1 \]
Possible pairs: \( (-4, 1), (-4, -1), (4, 1), (4, -1) \)

2. **For \( a = -3 \) or \( a = 3 \)**:
\[ 2(3^2) + 3b^2 = 35 \implies 18 + 3b^2 = 35 \implies 3b^2 = 17 \implies b^2 = \frac{17}{3} \quad \text{(not an integer)} \]

3. **For \( a = -2 \) or \( a = 2 \)**:
\[ 2(2^2) + 3b^2 = 35 \implies 8 + 3b^2 = 35 \implies 3b^2 = 27 \implies b^2 = 9 \implies b = \pm 3 \]
Possible pairs: \( (-2, 3), (-2, -3), (2, 3), (2, -3) \)

4. **For \( a = -1 \) or \( a = 1 \)**:
\[ 2(1^2) + 3b^2 = 35 \implies 2 + 3b^2 = 35 \implies 3b^2 = 33 \implies b^2 = 11 \quad \text{(not an integer)} \]

5. **For \( a = 0 \)**:
\[ 2(0^2) + 3b^2 = 35 \implies 3b^2 = 35 \implies b^2 = \frac{35}{3} \quad \text{(not an integer)} \]

6. **For \( a = -2 \) or \( a = 2 \)** (already calculated).

### Step 5: Collect all valid pairs
The valid integer pairs \( (a, b) \) we found are:
1. \( (-4, 1) \)
2. \( (-4, -1) \)
3. \( (4, 1) \)
4. \( (4, -1) \)
5. \( (-2, 3) \)
6. \( (-2, -3) \)
7. \( (2, 3) \)
8. \( (2, -3) \)

### Conclusion
Thus, the total number of valid pairs \( (a, b) \) is 8.

### Final Answer
The answer to the question is:
**Option C: 8**