1- We need to show that the 2 scalar softmax outputs of binary logistic regression are not necessary, and that you can make the exact same predictions using a single sigmoid output.
In a binary logistic regression classifier, you have two scalar outputs: $S(x) = [S(x)_0,\hspace{0.2cm}S(x)_1]$, where $S$ is the softmax function. The prediction would then be the class0 if $S(x)_0 > S(x)_1$, and class1 if $S(x)_0 < S(x)_1$. Since we know that $S(x)_0 + S(x)_1 = 1$ (property of softmax), then we can say that the prediction would be class0 if $S(x)_0 > 0.5$ and class1 if $S(x)_0 < 0.5$ (you can check that this is equivalent to the previous statement).
Already you can see that we do not actually need $S(X)_1$ to make a prediction, but we can make a prediction having only the value of $S(X)_0$. Now let's see how we can calculate the value of $S(X)_0$ using sigmoid:
$$
S(x)_0 = \dfrac{e^{x_0}}{e^{x_0}+e^{x_1}}= \dfrac{1}{1+\frac{e^{x_1}}{e^{x_0}}}= \dfrac{1}{1+e^{-(x_0-x_1)}}=\sigma(z)\quad\text{(where $z=x_0-x_1$)}
$$
$$z=x_0-x_1=(w_0-w_1)a+(b_0-b_1)=w'a+b'$$
So all you need to do is use a single node with the weight vector $w'=w_0-w_1$ and the bias $b'=b_0-b_1$ to calculate $z$ (where $w_0$ and $w_1$ are the weights corresponding to the two nodes of the binary logistic regression model), and then proceed to take the sigmoid of $z$.
2- Suppose a 3 layer neural network without activation functions:
$$
A_1 = W_1X\\
A_2 = W_2A_1\\
A_3 = W_3A_2
$$
$X$ is our input matrix where each column contains an input sample, $W_i$ is the weight matrix, and $A_i$ is the output at the $i$'th layer. Then we can simply write $A_3$ as:
$$
A_3 = W_{3}(A_2) = W_{3}(W_{2}A_1)= W_{3}(W_{2}(W_1X))=(W_{3}W_{2}W_1)X= W'X
$$
The output ($W'X$) is just a linear transformation ($W'$) applied to your inputs ($X$).
