From pp.104-105:
(1) [Second bullet point] How do we know that $H_A(\mathbb Z)=\{\text{group homomorphisms } \mathbb Z\to A\}$ is the underlying set of $A$? I can't even imagine why this could be true. $A$ was chosen to be an arbitrary group, and the set $H_A(\mathbb Z)$ is constructed using $A$.
(2) [Third bullet point] I think the result they are referring to is the fact that there is a bijection $$\{\text{group homomorphisms } \mathbb Z_p\to G\}\leftrightarrow\{\text{elements of } G \text{ of order }1 \text{ or } p\}$$ (this is not quite Example 4.1.5, but I used this fact to prove what is claimed in that example). This result should imply that for any $p$, $A$ and $A'$ have the same number of elements of order dividing $p$. Why does it follow that they have the same number of elements of order $p$?
For (1) notice that every homomorphism $\phi:\mathbb{Z}\rightarrow A$ is completely determined by its image $\phi(1)$.
For (2), in example 4.1.5 they define $U_p:\textbf{Grp}\rightarrow\textbf{Set}$ that maps $G$ to the set of elements of order 1 or of prime order in $G$ and in exercise 4.1.28 it is proven that $U_p\simeq Hom(\mathbb{Z}/p\mathbb{Z},-)$, so what is said follows right away (since the number of elements of order p is equal to $|U_p(G)|-1$).