Here is an answer to the first point and shows how you can also do the third point.

Let $\displaystyle \bigcup_{i=1}^{n+1} S_i=\{b_{1},b_{2},\ldots, b_{m}\}$. We consider the following table with rows indexed by the sets $S_1,S_2,\ldots, S_{n+1}$ and columns indexed by the elements $b_1,b_2,\ldots, b_m$. We put a $1$ in the cell $(i,j)$ with $1\leq i\leq k$ and $1\leq j\leq m$ if $b_j$ is an element of the set $S_i$. Let $x_i$ the number of $1$'s in the column $i$, or equivalently it is the number of sets in which the element $b_i$ appears with $1\leq i\leq m$.

Now we have that $S=\displaystyle \sum_{i=1}^{m} x_i$ is equal to the sum of the numbers in the whole table, and doing the sum by rows we obtain that $\displaystyle \sum_{i=1}^{n+1} x_i=\sum_{i=1}^{k} |A_i|=n(n+1)$.

Let us look further at $\displaystyle \sum_{1\leq i<j\leq n+1} |A_i\cap A_j|$. This is equal to the number of pairs of $1$'s who are in the same column. But this is equivalent in our notation to the fact $\displaystyle \sum_{1\leq i<j\leq n+1} |A_i\cap A_j|=\sum_{i=1}^{m} \dbinom{x_i}{2}$.

Now just use the convexity of the function $\binom{x}{2}$ to get that $$\sum_{i=1}^{m} \dbinom{x_i}{2}\geq m \dbinom{\frac{n(n+1)}{m}}{2}=\cfrac{(n(n+1)-m)n(n+1)}{2m}$$.

Since there are $\dbinom{n+1}{2}$ intersection the maximum size of the intersection is at least $\cfrac{n(n+1)-m}{m}$. Finally $m\leq 2n$ and you get the desired size almost namely $\cfrac{n-1}{2}$.

Here is an answer to the first point and shows how you can also do the third point.

Let $\displaystyle \bigcup_{i=1}^{n+1} S_i=\{b_{1},b_{2},\ldots, b_{m}\}$. We consider the following table with rows indexed by the sets $S_1,S_2,\ldots, S_{n+1}$ and columns indexed by the elements $b_1,b_2,\ldots, b_m$. We put a $1$ in the cell $(i,j)$ with $1\leq i\leq k$ and $1\leq j\leq m$ if $b_j$ is an element of the set $S_i$. Let $x_i$ the number of $1$'s in the column $i$, or equivalently it is the number of sets in which the element $b_i$ appears with $1\leq i\leq m$.

Now we have that $S=\displaystyle \sum_{i=1}^{m} x_i$ is equal to the sum of the numbers in the whole table, and doing the sum by rows we obtain that $\displaystyle \sum_{i=1}^{m} x_i=\sum_{i=1}^{n+1} |S_i|=n(n+1)$.

Let us look further at $\displaystyle \sum_{1\leq i<j\leq n+1} |S_i\cap S_j|$. This is equal to the number of pairs of $1$'s who are in the same column. But this is equivalent in our notation to the fact $\displaystyle \sum_{1\leq i<j\leq n+1} |S_i\cap S_j|=\sum_{i=1}^{m} \dbinom{x_i}{2}$.

Now just use the convexity of the function $\binom{x}{2}$ to get that $$\sum_{i=1}^{m} \dbinom{x_i}{2}\geq m \dbinom{\frac{n(n+1)}{m}}{2}=\cfrac{(n(n+1)-m)n(n+1)}{2m}$$.

Since there are $\dbinom{n+1}{2}$ intersection the maximum size of the intersection is at least $\cfrac{n(n+1)-m}{m}$. Finally $m\leq 2n$ and you get the desired size almost namely $\cfrac{n-1}{2}$.