`bb"Statement I"` The period of `f(x)=2cos""1/3(x-pi)+4sin""1/3(x-pi) " is " 3pi`.
`bb"Statement II"` If T is the period of f(x), then the period of f(ax+b) is `T/abs(a)`.

Answer

To solve the problem, we need to evaluate the two statements provided regarding the function \( f(x) = 2 \cos\left(\frac{1}{3}(x - \pi)\right) + 4 \sin\left(\frac{1}{3}(x - \pi)\right) \).

### Step 1: Determine the period of the cosine and sine components

The period of the cosine function \( \cos(kx) \) is given by:
\[ T_{\cos} = \frac{2\pi}{|k|} \]
For the term \( \cos\left(\frac{1}{3}(x - \pi)\right) \), we identify \( k = \frac{1}{3} \). Thus, the period is:
\[ T_{\cos} = \frac{2\pi}{\left|\frac{1}{3}\right|} = 2\pi \times 3 = 6\pi \]

Similarly, for the sine function \( \sin(kx) \), the period is also given by:
\[ T_{\sin} = \frac{2\pi}{|k|} \]
For the term \( \sin\left(\frac{1}{3}(x - \pi)\right) \), we again have \( k = \frac{1}{3} \). Thus, the period is:
\[ T_{\sin} = \frac{2\pi}{\left|\frac{1}{3}\right|} = 2\pi \times 3 = 6\pi \]

### Step 2: Determine the overall period of the function \( f(x) \)

The overall period of the function \( f(x) \) is the least common multiple (LCM) of the periods of its components:
\[ T_f = \text{lcm}(T_{\cos}, T_{\sin}) = \text{lcm}(6\pi, 6\pi) = 6\pi \]

### Step 3: Evaluate Statement I

Statement I claims that the period of \( f(x) \) is \( 3\pi \). Since we calculated the period to be \( 6\pi \), Statement I is **false**.

### Step 4: Evaluate Statement II

Statement II states that if \( T \) is the period of \( f(x) \), then the period of \( f(ax + b) \) is \( \frac{T}{|a|} \).

Given that the period \( T = 6\pi \), we apply the formula:
\[ \text{Period of } f(ax + b) = \frac{T}{|a|} = \frac{6\pi}{|a|} \]
This statement is generally true for periodic functions when transformed by a linear function \( ax + b \).

### Conclusion

- **Statement I** is **false**.
- **Statement II** is **true**.