1. Let's first write the expression clearly: $$\sum_{k=1}^3 \frac{\sin\left(\frac{k\pi}{12}\right) + \sin\left(\frac{(6-k)\pi}{12}\right)}{\cos\left(\frac{k\pi}{12}\right) + \cos\left(\frac{(6-k)\pi}{12}\right)} - \prod_{m=1}^2 \frac{\tan\left(\frac{\pi}{4} + \frac{m\pi}{12}\right) - \tan\left(\frac{\pi}{4} - \frac{m\pi}{12}\right)}{1 + \tan\left(\frac{\pi}{4} + \frac{m\pi}{12}\right)\tan\left(\frac{\pi}{4} - \frac{m\pi}{12}\right)}$$ 2. Evaluate the summation first. Recall the sum-to-product identities: $$\sin A + \sin B = 2 \sin\left( \frac{A+B}{2} \right) \cos\left( \frac{A-B}{2} \right)$$ $$\cos A + \cos B = 2 \cos\left( \frac{A+B}{2} \right) \cos\left( \frac{A-B}{2} \right)$$ Let $$A = \frac{k\pi}{12}$$ and $$B = \frac{(6-k)\pi}{12}$$. Calculate $$A+B = \frac{k\pi}{12} + \frac{(6-k)\pi}{12} = \frac{6\pi}{12} = \frac{\pi}{2}$$. Calculate $$A-B = \frac{k\pi}{12} - \frac{(6-k)\pi}{12} = \frac{(2k - 6)\pi}{12} = \frac{(k-3)\pi}{6}$$. Therefore, $$\sin A + \sin B = 2 \sin\left(\frac{\pi/2}{2}\right) \cos\left( \frac{(k-3)\pi/6}{2} \right) = 2 \sin\left( \frac{\pi}{4} \right) \cos\left( \frac{(k-3)\pi}{12} \right)$$ and $$\cos A + \cos B = 2 \cos\left(\frac{\pi/2}{2}\right) \cos\left( \frac{(k-3)\pi}{12} \right) = 2 \cos\left( \frac{\pi}{4} \right) \cos\left( \frac{(k-3)\pi}{12} \right)$$ 3. Simplify the fraction inside the summation: $$\frac{\sin A + \sin B}{\cos A + \cos B} = \frac{2 \sin(\frac{\pi}{4}) \cos(\frac{(k-3)\pi}{12})}{2 \cos(\frac{\pi}{4}) \cos(\frac{(k-3)\pi}{12})} = \frac{\sin(\frac{\pi}{4})}{\cos(\frac{\pi}{4})} = \tan\left(\frac{\pi}{4}\right) = 1$$ Note that this holds as long as $$\cos\left( \frac{(k-3)\pi}{12} \right) \neq 0$$ (which is true). 4. The summation thus becomes: $$\sum_{k=1}^3 1 = 3$$ 5. Next, evaluate the product: $$\prod_{m=1}^2 \frac{\tan\left( \frac{\pi}{4} + \frac{m\pi}{12} \right) - \tan\left( \frac{\pi}{4} - \frac{m\pi}{12} \right)}{1 + \tan\left( \frac{\pi}{4} + \frac{m\pi}{12} \right) \tan\left( \frac{\pi}{4} - \frac{m\pi}{12} \right)}$$ 6. Note that: $$\tan a - \tan b = \frac{\sin(a-b)}{\cos a \cos b}$$ and $$1 + \tan a \tan b = \frac{\cos(a-b)}{\cos a \cos b}$$ Therefore, $$\frac{\tan a - \tan b}{1 + \tan a \tan b} = \frac{\sin(a-b)}{\cos (a-b)} = \tan(a-b)$$ Here, $$a = \frac{\pi}{4} + \frac{m\pi}{12}, \quad b = \frac{\pi}{4} - \frac{m\pi}{12}$$ So, $$a - b = \left( \frac{\pi}{4} + \frac{m\pi}{12} \right) - \left( \frac{\pi}{4} - \frac{m\pi}{12} \right) = \frac{2 m \pi}{12} = \frac{m \pi}{6}$$ Hence, $$\frac{\tan\left( \frac{\pi}{4} + \frac{m\pi}{12} \right) - \tan\left( \frac{\pi}{4} - \frac{m\pi}{12} \right)}{1 + \tan\left( \frac{\pi}{4} + \frac{m\pi}{12} \right) \tan\left( \frac{\pi}{4} - \frac{m\pi}{12} \right)} = \tan\left( \frac{m \pi}{6} \right)$$ 7. Therefore, the product simplifies to: $$\prod_{m=1}^2 \tan\left( \frac{m\pi}{6} \right) = \tan\left( \frac{\pi}{6} \right) \times \tan\left( \frac{2\pi}{6} \right) = \tan\left( \frac{\pi}{6} \right) \times \tan\left( \frac{\pi}{3} \right)$$ Recall: $$\tan\left( \frac{\pi}{6} \right) = \frac{1}{\sqrt{3}}$$ $$\tan\left( \frac{\pi}{3} \right) = \sqrt{3}$$ 8. Multiply: $$\frac{1}{\sqrt{3}} \times \sqrt{3} = 1$$ 9. Finally, the entire expression equals: $$3 - 1 = 2$$ **Answer:** $$\boxed{2}$$