$1)$. Prove a graph with $2n$ vertices and $n+2$ edges has an independent set of size $n$ (there are $n$ vertices such that no two of them are adjacent ).
$2)$.Find the number of graphs with $2n$ vertices and $n+3$ edges , such that among any $n$ vertices there is an edge connecting two of them
### Part 1:
We need to prove that a graph \( G \) with \( 2n \) vertices and \( n+2 \) edges has an independent set of size \( n \). 

1. **Complement Graph**: Consider the complement graph \( G' \) of \( G \). In \( G' \), an edge exists between two vertices if and only if it does not exist in \( G \). 

2. **Vertices and Edges in \( G' \)**: Since \( G \) has \( 2n \) vertices and \( n+2 \) edges, \( G' \) will have \( 2n \) vertices and \( \binom{2n}{2} - (n+2) \) edges. 
   \[
   |E(G')| = \binom{2n}{2} - (n+2) = \frac{2n(2n-1)}{2} - (n+2) = 2n^2 - n - 2
   \]

3. **Finding a \( K_n \) in \( G' \)**: We need to show that \( G' \) contains a complete subgraph \( K_n \). According to the Mantel-Turán Theorem, a graph with \( 2n \) vertices and more than \( \frac{1}{2} \cdot \frac{n-2}{n-1} \cdot (2n)^2 \) edges contains a \( K_n \).

4. **Applying Mantel-Turán Theorem**:
   \[
   2n^2 - n - 2 > \frac{1}{2} \cdot \frac{n-2}{n-1} \cdot (2n)^2
   \]
   Simplifying the right-hand side:
   \[
   \frac{1}{2} \cdot \frac{n-2}{n-1} \cdot 4n^2 = 2n^2 \cdot \frac{n-2}{n-1}
   \]
   We need to show:
   \[
   2n^2 - n - 2 > 2n^2 \cdot \frac{n-2}{n-1}
   \]
   Simplifying further:
   \[
   2n^2 - n - 2 > 2n^2 - \frac{4n^2}{n-1}
   \]
   This inequality holds true for \( n \geq 2 \).

5. **Conclusion**: Since \( G' \) contains a \( K_n \), \( G \) must have an independent set of size \( n \).

### Part 2:
We need to find the number of graphs with \( 2n \) vertices and \( n+3 \) edges such that among any \( n \) vertices, there is an edge connecting two of them.

1. **Complement Graph**: Consider the complement graph \( G' \) of \( G \). \( G' \) will have \( 2n \) vertices and \( \binom{2n}{2} - (n+3) \) edges.
   \[
   |E(G')| = \binom{2n}{2} - (n+3) = 2n^2 - n - 3
   \]

2. **Applying Mantel-Turán Theorem**: The graphs that follow the condition must have the complement \( G' \) satisfying the Turán condition:
   \[
   2n^2 - n - 3 \leq \frac{1}{2} \cdot \frac{n-2}{n-1} \cdot (2n)^2
   \]
   Simplifying the right-hand side:
   \[
   \frac{1}{2} \cdot \frac{n-2}{n-1} \cdot 4n^2 = 2n^2 \cdot \frac{n-2}{n-1}
   \]
   We need to show:
   \[
   2n^2 - n - 3 \leq 2n^2 - \frac{4n^2}{n-1}
   \]
   Simplifying further:
   \[
   2n^2 - n - 3 \leq 2n^2 - 2n - 2
   \]
   This inequality holds true for \( n \geq 3 \).

3. **Number of Graphs**: The number of graphs is the number of ways to partition the set of \( 2n \) elements into 2 sets of 3 elements and \( n-3 \) sets of 2 elements:
   \[
   \frac{(2n)!}{9 \cdot 2^{n-3} \cdot (n-3)! \cdot 2!}
   \]

The final answer is \( \boxed{ \frac{(2n)!}{9 \cdot 2^{n-3} \cdot (n-3)! \cdot 2!} } \)