Recall the tangent addition and subtraction formulas.

The formulas for $\tan(A+B)$ and $\tan(A-B)$ are:

$$\tan(A+B) = \frac{\tan A + \tan B}{1 - \tan A \tan B}$$ $$\tan(A-B) = \frac{\tan A - \tan B}{1 + \tan A \tan B}$$

We know that $\tan\left(\frac{\pi}{4}\right) = 1$.

$$\tan\left(\frac{\pi}{4} + \theta\right) = \frac{\tan\left(\frac{\pi}{4}\right) + \tan\theta}{1 - \tan\left(\frac{\pi}{4}\right)\tan\theta} = \frac{1 + \tan\theta}{1 - \tan\theta}$$ $$\tan\left(\frac{\pi}{4} - \theta\right) = \frac{\tan\left(\frac{\pi}{4}\right) - \tan\theta}{1 + \tan\left(\frac{\pi}{4}\right)\tan\theta} = \frac{1 - \tan\theta}{1 + \tan\theta}$$

$$\cot\left(\frac{\pi}{4} + \theta\right) = \frac{1}{\tan\left(\frac{\pi}{4} + \theta\right)} = \frac{1 - \tan\theta}{1 + \tan\theta}$$ $$\cot\left(\frac{\pi}{4} - \theta\right) = \frac{1}{\tan\left(\frac{\pi}{4} - \theta\right)} = \frac{1 + \tan\theta}{1 - \tan\theta}$$

$$\cot\left(\frac{\pi}{4} + \theta\right)\cot\left(\frac{\pi}{4} - \theta\right) = \left(\frac{1 - \tan\theta}{1 + \tan\theta}\right) \times \left(\frac{1 + \tan\theta}{1 - \tan\theta}\right)$$

The terms $(1 - \tan\theta)$ and $(1 + \tan\theta)$ cancel out.

$$\cot\left(\frac{\pi}{4} + \theta\right)\cot\left(\frac{\pi}{4} - \theta\right) = 1$$

The final answer is $\boxed{\text{1}}$.
The correct option is (3).