elements in a set

a) Note that 

\[(1)             9<k^2<40  \Leftrightarrow  3<|k|<\sqrt{40}=6.32.\]

Also $k\in \mathbb{Z}$ means that $k$ must be a positive innegative integer. The set of integers that satisfy (1) is 

\[\{k\in \mathbb{Z}:9<k^2<40\}=\{4,5,6,-4,-5,-6\}.\]

(b) Note that

\[\{k\in \mathbb{Z}:-4<k^2<16\}, \frac{k}{3}\in \mathbb{N}\}\]
is the set of integers between $-4$ and $16$ such that $k/3$ is a postive integer. So 

\[\{k\in \mathbb{Z}:-4<k^2<16\}, \frac{k}{3}\in \mathbb{N}\}=\{3,6,9,15\}.\]

Also 
\[\{k\in \mathbb{Z}: k | 12\}=\{1,-1,2,-2,3,-3,4,-4,6,-6,12,-12\},\]
which is the set of integers (positive, negative, or zero) which are factors of $12$. So 
\[\{k\in \mathbb{Z}:-4<k^2<16\}, \frac{k}{3}\in \mathbb{N}\} \cup \{k\in \mathbb{Z}: k | 12\}\]
\[=\{k\in \mathbb{Z}: k | 12\}=\{1,-1,2,-2,3,-3,4,-4,6,-6,12,-12\}\cup \{3,6,9,15\}\]
\[=\{1,-1,2,-2,3,-3,4,-4,6,-6,12,-12, 9, 15\}.     (2)\]

So 
\[|\{k\in \mathbb{Z}:-4<k^2<16\}, \frac{k}{3}\in \mathbb{N}\} \cup \{k\in \mathbb{Z}: k | 12\}|=14,\]
which is the number of elements in the set (2).

Note that $\mathbb{Z}$ denotes the set of integers (could be positive, negative, or zero), but $\mathbb{N}$ is the set of positive integers.