The first question has a very well known solution (try searching for "birthday paradox"), so I will go straight to the second one.
For the second question: Suppose we label the people $p_1,\dots,p_n$ and begin by imposing that $p_1$ has its birthday on 6th September (let's call $B$ the set of people having their birthday on 6th September and write $p_1\in B$). Then impose that $p_2$ has its birthday on 6th September too. This has probability $\frac{1}{365}\cdot\frac{1}{365}$.
The next case to consider is (while keeping $p_1$'s birthday on 6th September) that $p_3$ has its birthday on 6th September and $p_2$ does not have its birthday on 6th September, which has probability $\frac{1}{365}\cdot\frac{364}{365}\cdot\frac{1}{365}$. The case after this is to exclude $p_3$ as well and set $p_1\in B, p_2\notin B, p_3\notin B, p_4\in B$ which has probability $\frac{1}{365}\cdot\frac{364}{365}\cdot\frac{364}{365}\cdot\frac{1}{365}$, and so on.
It is necessary to exclude $p_2\in B$ to avoid counting twice cases like $B=\{p_1,p_2,p_3\}$ for instance, which are taken into account in the first case. Like this, the probability of two people having birthday on 6th September and one of them being $p_1$ is
\begin{equation}\tag{1}
\frac{1}{365}\cdot\frac{1}{365}+\frac{1}{365}\cdot\frac{364}{365}\cdot\frac{1}{365}+\ldots+\frac{1}{365}\cdot\frac{364}{365}\cdot\stackrel{n-2}{\ldots}\cdot\frac{364}{365}\cdot\frac{1}{365}
\end{equation}
where in the last sum there are $n-2$ of the $\frac{364}{365}$ fractions.
This can be rewritten as
\begin{equation}\tag{2}
\bigg(\frac{1}{365}\bigg)^2\bigg(1+\frac{364}{365}+\ldots+\big(\frac{364}{365}\big)^{n-2}\bigg)
\end{equation}
which in turn can be transformed into
\begin{equation}\tag{3}
\bigg(\frac{1}{365}\bigg)^2\bigg(\frac{1-\big(\frac{364}{365}\big)^{n-1}}{1-\frac{364}{365}}\bigg)=\frac{1}{365}\bigg(1-\big(\frac{364}{365}\big)^{n-1}\bigg)
\end{equation}
using the geometric progression formula.
Now that we dealt with all the cases involving $p_1\in B$, we impose $p_1\notin B$. We start by counting the cases where $p_2\in B$: first $p_2,p_3\in B$; then $p_2\in B, p_3\notin B, p_4\in B$ and so on.
The probability of this has an expression similar to (1), except that there is a $\frac{364}{365}$ at the beginning of each term (due to the condition $p_1\notin B$) and there is one term less because there is one person less to worry about (namely $p_1$):
\begin{equation}\tag{1'}
\frac{364}{365}\bigg(\frac{1}{365}\cdot\frac{1}{365}+\frac{1}{365}\cdot\frac{364}{365}\cdot\frac{1}{365}+\ldots+\frac{1}{365}\cdot\frac{364}{365}\cdot\stackrel{(n-2)-1=n-3}{\ldots}\cdot\frac{364}{365}\cdot\frac{1}{365}\bigg)
\end{equation}
Accordingly, (1') is equivalent to
\begin{equation}\tag{2'}
\bigg(\frac{1}{365}\bigg)^2\cdot\frac{364}{365}\bigg(1+\frac{364}{365}+\ldots+\big(\frac{364}{365}\big)^{n-3}\bigg)
\end{equation}
and
\begin{equation}\tag{3'}
\bigg(\frac{1}{365}\bigg)^2\cdot\frac{364}{365}\bigg(\frac{1-\big(\frac{364}{365}\big)^{n-2}}{1-\frac{364}{365}}\bigg)=\frac{1}{365}\cdot\frac{364}{365}\bigg(1-\big(\frac{364}{365}\big)^{n-2=(n-1)-1}\bigg)
\end{equation}
We can expand a bit more and get
\begin{equation}\tag{4'}
\frac{1}{365}\cdot\bigg(\frac{364}{365}-\big(\frac{364}{365}\big)^{n-1}\bigg)
\end{equation}
This process goes on until there are $n-2$ people excluded from $B$ and the only possibility is $p_1,\dots,p_{n-2}\notin B, p_{n-1},p_n\in B$. Here, the version of (1) is just
\begin{equation}\tag{1''}
\bigg(\frac{364}{365}\bigg)^{n-2}\bigg(\frac{1}{365}\cdot\frac{1}{365}\bigg)
\end{equation}
as there must be $(n-2)-(n-2)=0$ copies of the $\frac{364}{365}$ in the last term.
This (1'') is clearly equivalent to
\begin{equation}\tag{4''}
\frac{1}{365}\cdot\bigg(\big(\frac{364}{365}\big)^{n-2}-\big(\frac{364}{365}\big)^{n-1}\bigg)
\end{equation}
Now, to compute the total probability two people being in $B$ we do $(3) + (4') + \dots + (4'')$, which yields
\begin{equation}\tag{5}
\frac{1}{365}\cdot\bigg(1-\big(\frac{364}{365}\big)^{n-1}+\frac{364}{365}-\big(\frac{364}{365}\big)^{n-1}+\dots+\big(\frac{364}{365}\big)^{n-2}-\big(\frac{364}{365}\big)^{n-1}\bigg)
\end{equation}
We see there are $(n-1)$ occurrences of $\big(\frac{364}{365}\big)^{n-1}$ and the other terms form a geometric progression for which we use the same formula as before. We get
\begin{equation}\tag{6}
\frac{1}{365}\cdot\bigg(\frac{1-\big(\frac{364}{365}\big)^{n-1}}{1-\frac{364}{365}}-(n-1)\big(\frac{364}{365}\big)^{n-1}\bigg)
\end{equation}
Expand
\begin{equation}\tag{7}
1-\big(\frac{364}{365}\big)^{n-1}-\frac{(n-1)}{365}\big(\frac{364}{365}\big)^{n-1}=1-\big(\frac{364}{365}\big)^{n-1}\bigg(1+\frac{(n-1)}{365}\bigg)
\end{equation}
Finally, impose (7) $\geq 0.5$, from where you can see that $n\geq 613$
