Item 3.
For variance, we will use an alternate form of the definition that is useful when looking for cancellation options with the numerous factorials in the hypergeometric probability function 6.4. Indeed, you can easily notice that
\begin{equation*}
\sigma^2 = E[X^2] - \mu^2 = E[X^2-X]+E[X] -\mu^2 = E[X(X-1)] + \mu - \mu^2.
\end{equation*}
Since we have \(\mu = r \frac{n_1}{n}\) from above then let’s focus on the first term only and use the substitutions
\begin{gather*}
y = x-2\\
n_3 = n_1-2\\
s = r-2\\
m = n-2
\end{gather*}
to get
\begin{gather*}
E[X(X-1)] = \sum_{x=0}^n x(x-1) \frac{\binom{n_1}{x} \binom{n-n_1}{r-x}}{\binom{n}{r}}\\
= \sum_{x=2}^n x(x-1) \frac{\frac{n_1!}{x(x-1)(x-2)!(n_1-x)! } \binom{n-n_1}{r-x}}{\binom{n}{r}}\\
= \sum_{x=2}^n \frac{\frac{n_1!}{(x-2)!(n_1-x)! } \frac{n_2!}{(r-x)!(n_2-r+x)!}}{\binom{n}{r}}\\
= n_1 \cdot (n_1-1) \cdot
\sum_{x=2}^n \frac{\frac{(n_3)!}{(x-2)!(n_3 -(x-2))! } \frac{n_2!}{((r-2)-(x-2))!(n_2-(r-2)+(x-2))!}}{\binom{n}{r}}\\
= n_1 \cdot (n_1-1) \sum_{y=0}^{m} \frac{\frac{(n_3)!}{y!(n_3 -y)! } \frac{n_2!}{(s-y)!(n_2-s+y)!}}{\binom{n}{r}}\\
= \frac{n_1 \cdot (n_1-1) \cdot r \cdot (r-1)}{n (n-1)} \sum_{y=0}^{m} \frac{\binom{(n_3)}{y} \binom{n_2}{s-y}}{\binom{m}{s}}\\
= \frac{n_1 \cdot (n_1-1) \cdot r \cdot (r-1)}{n (n-1)}
\end{gather*}
where we have used the summation formula above that showed that f(x) was a probability function.
Putting this together with the earlier formula gives
\begin{equation*}
\sigma^2 = \frac{n_1 \cdot (n_1-1) \cdot r \cdot (r-1)}{n (n-1)} + r \frac{n_1}{n} - \left ( r \frac{n_1}{n} \right )^2.
\end{equation*}