(a)
To prove that $s^2$ is unbiased in estimating $\sigma^2$, we need to show that $\mathrm{E}(s^2) = \sigma^2$.
To Prove that we shall first prove the following Lemmas:
Lemma 1
$\mathrm{E}(X_i) = \mu$, where $\mu$ denotes the population mean.
Proof
Since this is a random sampling with replacement and every element having equal chance to be selected:
$$
P(X_i = x_j) = \frac 1 N
$$
Thus $\mathrm{E}(X_i) = \sum_{i=1}^{i=n} x_i P(X_i = x_j) = \frac 1 N \sum_{i=1}^{i=n} x_i$
Thus $\mathrm{E}(X_i) = \mu$
Lemma 2
$$
\mathrm{E}(\bar X) = \mu
$$
Proof
$$
\mathrm{E}(\bar X) = \mathrm{E} \left( \frac 1 N \sum_{i=1}^{i=n} X_i \right) = \frac 1 n \sum_{i=1}^{i=n} \mathrm{E} (X_i) = \frac {n \mu} {n} = \mu
$$
Now, we will prove the main result $\mathrm{E}(s^2) = \sigma^2$
We have:
$$
\,
\begin{align*}
s^2 &= \frac 1 {n-1} \sum_{i=1}^{i=n} { \left( X_i - { \bar X} \right) }^2 \\
\Rightarrow (n-1) s^2 &= \sum_{i=1}^{i=n} \left( X_i^2 + {\bar X}^2 - 2 {X_i} { \bar X} \right) \\
&= \sum^{n}_{i=1} X_i^2 + \sum^{n}_{i=1} {\bar X}^2 - 2 {\bar X} \sum^{n}_{i=1} X_i \\
&= \sum^{n}_{i=1} X_i^2 \quad+\; n {\bar X}^2 - 2 {\bar X} n {\bar X} && \text{Using Lemma 1} \\
&= \sum^{n}_{i=1} X_i^2 \quad- n {\bar X}^2 \\
(n-1)\mathrm{E}(s^2) &= \sum^{n}_{i=1} \mathrm{E}(X_i^2) \quad- n\mathrm{E} ({\bar X}^2) && \text{Taking Expectation, } \\
&= \sum^{n}_{i=1} \left( \mathrm{Var}(X_i) + (\mathrm{E}X_i)^2 \right) \quad- n\mathrm{E} ({\bar X}^2) && \text{Using $\mathrm{Var}(X_i) = \mathrm{E}X_i^2 - (\mathrm{E}X_i)^2$} \\
&= n\mathrm{Var}{X_1} + \sum^{n}_{i=1}(\mathrm{E}X_i)^2 \quad-\; n\mathrm{E} {(\bar X}^2) && \text{ Since all $X_i$ are i.i.d., $ \mathrm{Var}(X_i) = \mathrm{Var}(X_1)$} \\
&= n \sigma^2 + \sum^{n}_{i=1} \left( (\mathrm{E}X_i)^2 - \mathrm{E} ({\bar X}^2) \right) \\
&= n \sigma^2 - \sum^{n}_{i=1} \left( \mathrm{E}({\bar X}^2) - (\mathrm{E}X_i)^2 \right) && \text{ Rearranging terms } \\
&= n \sigma^2 - \sum^{n}_{i=1} \left( \mathrm{E}({\bar X}^2) - (\mathrm{E}\bar X)^2 \right) && \text{ Using Lemma 2, $\mathrm{E} X_i = \mathrm{E}{\bar X} = \mu$ } \\
&= n { \sigma }^2 - \sum^{n}_{i=1} \mathrm{ Var } \left( \bar X \right) && \text{ Using Variance formulae } \\
&= n { \sigma }^2 - n \frac{\sigma^2}{n} && \text{ Since $\mathrm{Var}(\bar X) = \frac {\sigma^2} n$, as shown in book, Pg-207 } \\
&= \left( n-1 \right) \sigma^2
\end{align*}
\,
$$
Cancelling $n-1$ from both sides, it follows $\mathrm{E}(s^2) = \sigma^2$.
(b)
No, it is not an unbiased estimate for sigma. By Jensen’s Inequality:
$$
\,
\begin{align*}
\mathrm{E}(s) &= \mathrm{E} \left(\sqrt s^2 \right) \\
&\leq \sqrt { \mathrm{E} ( s^2 ) } && \text{ Using Jensen's inequality } \\
&= \sqrt{ \sigma^2 } && \text{ From part-a } \\
&= \sigma
\end{align*}
\,
$$
Thus $\, \mathrm{E}(s) \leq \sigma \,$. It follows that $\, \mathrm{E}(s) \,$ is not always equal to $\sigma$, or $s$ is not an unbiased estimate of $\sigma$.
(c)
We need to show $\, \mathrm{E} \left( \frac{s^2}{n} \right) = \sigma^2_{\bar X} \,$
We have:
$$
\,
\begin{align*}
\mathrm{E} \left( \frac{s^2}{n} \right) &= \frac 1 n \mathrm{E} (s^2) \\
&= \frac 1 n \sigma^2 && \text{ From Part a } \\
&= \sigma^2_{\bar X} && \text{ As shown in book, Pg-207 }
\end{align*}
\,
$$
(d)
We have:
$$
\,
\begin{align*}
\mathrm{E} \left( \frac{N^2 s^2}{n} \right) &= N^2 \mathrm{E} \frac {s^2} n \\
&= N^2 \sigma^2_{\bar X} && \text{ From part b } \\
&= N^2 \mathrm{Var} (\bar X) \\
&= \mathrm{Var} (N \bar X) \\
&= \mathrm{Var} (T) \\
&= \sigma^2_T
\end{align*}
\,
$$
(e)
To show $\, \mathrm{E} \left( \frac { \hat p (1- \hat p) } {n-1} \right) = \sigma^2_{\hat p} \,$, where $\, \hat p = \bar X = \frac 1 n \sum_{i=1}^{n} X_i \,$, the sample proportion.
We have:
$$
\,
\begin{align*}
\sigma^2_{\bar p} &= \mathrm{Var} \bar X \\
&= \frac {\sigma^2} n && \text{Shown in book Pg 207, where $\sigma^2$ is population variance} \\
&= \frac 1 n \mathrm{E} s^2 && \text{From part a} \\
&= \frac 1 n \mathrm{E} \left( \frac 1 {n-1} \sum_{i=1}^{n} (X_i - \bar X)^2 \right) \\
&= \frac 1 {n(n-1)} \mathrm{E} \left( \sum_{i=1}^{n} X_i^2 \quad - n\bar X^2 \right) && \text{Shown in steps of part a}\\
&= \frac 1 {n(n-1)} \mathrm{E} \left( \sum_{i=1}^{n} X_i \quad - n\bar X^2 \right) && \text{Since $X_i$ is $0$ or $1$, thus $X_i^2 = X_i$}\\
&= \frac 1 {n(n-1)} \mathrm{E} ( n\hat p - n\hat p^2) \\
&= \frac 1 {(n-1)} \mathrm{E} ( \hat p - \hat p^2) \\
&= \mathrm{E} \left(\frac {\hat p(1 - \hat p)} {n-1} \right) \\
\end{align*}
\,
$$
$$\tag*{$\blacksquare$} $$