a) Calculate the product  $$\left(1+\frac{1}{2}\right)   \left(1+\frac{1}{3}\right)  \left(1+\frac{1}{4}\right)...  \left(1+\frac{1}{2006}\right)  \left(1+\frac{1}{2007}\right)$$
b) Let the set $$A =\left\{\frac{1}{2},  \frac{1}{3},\frac{1}{4}, ...,\frac{1}{2006}, \frac{1}{2007}\right\}$$
Determine the sum of all products of $2$, of $4$, of $6$,... , of $2004$ ¸and of $ 2006$ different elements of the set $A$.
### Part (a)
1. Consider the product:
   \[
   \left(1 + \frac{1}{2}\right) \left(1 + \frac{1}{3}\right) \left(1 + \frac{1}{4}\right) \cdots \left(1 + \frac{1}{2006}\right) \left(1 + \frac{1}{2007}\right)
   \]

2. Each term in the product can be rewritten as:
   \[
   1 + \frac{1}{k} = \frac{k+1}{k}
   \]
   Therefore, the product becomes:
   \[
   \prod_{k=2}^{2007} \left(1 + \frac{1}{k}\right) = \prod_{k=2}^{2007} \frac{k+1}{k}
   \]

3. Notice that this is a telescoping product:
   \[
   \frac{3}{2} \cdot \frac{4}{3} \cdot \frac{5}{4} \cdots \frac{2008}{2007}
   \]

4. All intermediate terms cancel out, leaving:
   \[
   \frac{2008}{2}
   \]

5. Simplifying the fraction:
   \[
   \frac{2008}{2} = 1004
   \]

Thus, the product is:
\[
\boxed{1004}
\]

### Part (b)
1. Let the set \( A = \left\{ \frac{1}{2}, \frac{1}{3}, \frac{1}{4}, \ldots, \frac{1}{2006}, \frac{1}{2007} \right\} \).

2. We need to determine the sum of all products of 2, 4, 6, ..., 2004, and 2006 different elements of the set \( A \).

3. Consider the polynomial:
   \[
   P(x) = \prod_{k=2}^{2007} \left(1 + \frac{x}{k}\right)
   \]

4. The sum of the products of \( m \) different elements of \( A \) corresponds to the coefficient of \( x^m \) in the expansion of \( P(x) \).

5. Since \( P(x) \) is a polynomial of degree 2006, the sum of the products of all even numbers of elements (2, 4, 6, ..., 2004, 2006) is given by:
   \[
   P(1) - 1
   \]

6. We already know from part (a) that:
   \[
   P(1) = \prod_{k=2}^{2007} \left(1 + \frac{1}{k}\right) = 1004
   \]

7. Therefore, the sum of all products of 2, 4, 6, ..., 2004, and 2006 different elements of the set \( A \) is:
   \[
   1004 - 1 = 1003
   \]

Thus, the sum is:
\[
\boxed{1003}
\]