Partitioning evens as sum of evens

Take the set $$\{a_1,a_2,a_3,a_4,a_5,a_6,a_7,a_8\}$$.

We can partition according to rules.

1. Every member in the partition has even number of elements.

2. Every member in partition have to be consecutive.

For example partitions above are:

1. $$\{(a_1,a_2),(a_3,a_4,a_5,a_6,a_7,a_8)\}$$.

2. $$\{(a_1,a_2),(a_3,a_4),(a_5,a_6,a_7,a_8)\}$$.

3. $$\{(a_1,a_2),(a_3,a_4),(a_5,a_6),(a_7,a_8)\}$$.

4. $$\{(a_1,a_2),(a_3,a_4,a_5,a_6),(a_7,a_8)\}$$.

5. $$\{(a_1,a_2,a_3,a_4,a_5,a_6),(a_7,a_8)\}$$.

6. $$\{(a_1,a_2,a_3,a_4),(a_5,a_6,a_7,a_8)\}$$.

7. $$\{(a_1,a_2,a_3,a_4,a_5,a_6,a_7,a_8)\}$$.

Essentially asking if we are given even number how many ways can write as sum of evens?

Here $$2+6=2+2+4=2+2+2+2=2+4+2=6+2=4+2+2=4+4=8$$.

In the example, imagine that there is a wall between $$a_2$$ and $$a_3$$, another between $$a_4$$ and $$a_5$$ and a third between $$a_6$$ and $$a_7.$$ Then you're just selecting which of the three walls to raise, so there are $$2^3=8$$ possibilities.
In you examples, $$1$$ corresponds to raising the first wall only, $$3$$ to raising all the walls, and $$7$$ to raising none of the walls. You have overlooked the partition $$(a_1,a_2,a_3,a_4)(a_5,a_6)(a_7,a_8.)$$
Think about partitioning this set as placing a wall between two elements. We can only place a wall after an element with an even index, i.e., after 2, 4, or 6, if we place a wall at all. The question then amounts to "how many subsets are there of the collection of possible walls $$\{2,4,6\}$$"? You might remember that there are $$2^n$$ subsets of any set with $$n$$ elements, so there are $$8$$ such partitions of $$8$$.
The only one you did not list was $$\{(a_1,a_2,a_3,a_4),(a_5,a_6),(a_7,a_8)\}$$.
In general, if instead of 8 elements you had $$2k$$, then there would be a possible wall for each even number less than $$2k$$, of which there are $$k-1$$. So there are $$2^{(n/2)-1}$$ such "partitions" of any even number.