Statement-1: `cos((pi)/(10))cos((2pi)/(10))cos((4pi)/(10))=(cos((8pi)/(10)))/(2^(3)sin((pi)/(10)))`
Statement-2: `cos A.cos2A.cos2^(2)A.......cos(2^(n-1)A)=(sin(2^(n)A))/(2^(n)sinA)`

Answer

To solve the problem, we need to analyze both statements and determine their validity step by step.

### Statement 1:
We need to verify the equation:
\[ \cos\left(\frac{\pi}{10}\right) \cos\left(\frac{2\pi}{10}\right) \cos\left(\frac{4\pi}{10}\right) = \frac{\cos\left(\frac{8\pi}{10}\right)}{2^3 \sin\left(\frac{\pi}{10}\right)} \]

**Step 1: Rewrite the Left-Hand Side (LHS)**

Start with the left-hand side:
\[ \cos\left(\frac{\pi}{10}\right) \cos\left(\frac{2\pi}{10}\right) \cos\left(\frac{4\pi}{10}\right) \]

**Hint:** Use the identity \( \sin(2a) = 2 \sin(a) \cos(a) \).

**Step 2: Apply the Identity**

We can multiply and divide by \( 2 \sin\left(\frac{\pi}{10}\right) \):
\[ = \frac{2 \sin\left(\frac{\pi}{10}\right) \cos\left(\frac{\pi}{10}\right) \cos\left(\frac{2\pi}{10}\right) \cos\left(\frac{4\pi}{10}\right)}{2 \sin\left(\frac{\pi}{10}\right)} \]

The numerator becomes:
\[ = \frac{\sin\left(\frac{2\pi}{10}\right) \cos\left(\frac{2\pi}{10}\right) \cos\left(\frac{4\pi}{10}\right)}{2 \sin\left(\frac{\pi}{10}\right)} \]

**Hint:** Recognize that \( \sin(2a) = 2 \sin(a) \cos(a) \).

**Step 3: Simplify Further**

Now, we can apply the identity again:
\[ = \frac{\sin\left(\frac{2\pi}{10}\right) \cdot \frac{1}{2} \sin\left(\frac{4\pi}{10}\right)}{2 \sin\left(\frac{\pi}{10}\right)} \]

Continuing this process leads to:
\[ = \frac{\sin\left(\frac{8\pi}{10}\right)}{2^3 \sin\left(\frac{\pi}{10}\right)} \]

**Step 4: Compare Both Sides**

Now we have:
\[ \cos\left(\frac{\pi}{10}\right) \cos\left(\frac{2\pi}{10}\right) \cos\left(\frac{4\pi}{10}\right) = \frac{\sin\left(\frac{8\pi}{10}\right)}{2^3 \sin\left(\frac{\pi}{10}\right)} \]

Since \( \sin\left(\frac{8\pi}{10}\right) = \cos\left(\frac{8\pi}{10}\right) \), we find that the statement is indeed true.

### Statement 2:
We need to verify the equation:
\[ \cos A \cos(2A) \cos(2^2 A) \ldots \cos(2^{n-1} A) = \frac{\sin(2^n A)}{2^n \sin A} \]

**Step 1: Rewrite the Left-Hand Side (LHS)**

Start with the left-hand side:
\[ \cos A \cos(2A) \cos(2^2 A) \ldots \cos(2^{n-1} A) \]

**Hint:** Again, use the identity \( \sin(2a) = 2 \sin(a) \cos(a) \).

**Step 2: Apply the Identity**

Multiply and divide by \( 2 \sin A \):
\[ = \frac{2 \sin A \cos A \cos(2A) \ldots \cos(2^{n-1} A)}{2 \sin A} \]

The numerator becomes:
\[ = \frac{\sin(2A) \cos(2A) \ldots \cos(2^{n-1} A)}{2 \sin A} \]

**Step 3: Continue Simplifying**

Continuing this process leads to:
\[ = \frac{\sin(2^n A)}{2^n \sin A} \]

**Step 4: Compare Both Sides**

Now we have:
\[ \cos A \cos(2A) \cos(2^2 A) \ldots \cos(2^{n-1} A) = \frac{\sin(2^n A)}{2^n \sin A} \]

This statement is also true.

### Conclusion:
- **Statement 1** is true.
- **Statement 2** is true.

### Final Answer:
Both statements are correct.