Let $G = (\mathbb Z, +)$ and $H,K ≤ G$ where $H = ⟨12⟩$ and $K =\mathbb⟨18⟩$.
a) Define $H\cap K$.
b) Define $H + K$.
a) Am I understanding correctly that this means that $H = \{x \in \mathbb Z : 18 | x\}$ and $K = \{x \in \mathbb Z : 12 | x\}$? So then would it be sufficient to say that the intersection is the elements of $K$ that are divisible by $12$ or the elements of $H$ that are divisible by $18$?
b) I'm unsure of the notation used here; are we summing individual elements?
While @Shaun is right that you should generally ask only one question at a time, I think these could be construed as related enough to be one post. Your questions seem to mostly be about what the problem is asking, so I'll answer those questions about terminology, hopefully leaving you in a better position to answer the question for yourself.
a) Correct. Given a group $G$ and a subset $S$ of $G$ (with $S$ not necessarily a group), we define $\left< S \right>$ to be the smallest subgroup of $G$ that contains $S$. One way to write this is $$\left<S \right> : = \bigcap_{\textrm{$L$ is a subgroup of $G$ and $S \subseteq L$}} L .$$ This is a group because the intersection of groups is a group. We typically refer to $\left< S \right>$ as the subgroup generated by $S$. But on a more intuitive level, you can think of $\left< S \right>$ as what happens when you extend $S$ to be closed under products and inverses. So we can equivalently write $$ \left< S \right> = \bigcup_{n \in \mathbb{N}} \left\{ s_1^{\epsilon_1} \cdots s_n^{\epsilon_n} : s_i \in S, \epsilon_i \in \{0, 1\} \textrm{ for all } i \in [1, n] \right\}. $$ When $g$ is a single element of $G$, we sometimes write $\left< g \right> : = \left< \{ g \} \right>$. In this case, we have $\left< g \right> = \left\{ g^k : k \in \mathbb{Z} \right\}$. In you particular situation, the group $G$ is abelian, so we often write the group operation as addition instead of multiplication. Thus \begin{align*} \left< S \right> & = \bigcup_{n \in \mathbb{N}} \left\{ a_1 s_1 + \cdots + a_n s_n : s_i \in S, a_i \in \mathbb{Z} \textrm{ for all } i \in [1, n] \right\}, \\ \left< g \right> & = \left\{ k g : k \in \mathbb{Z} \right\} . \end{align*} In the case of this problem specifically, remember that intersection is an "and," so if $H = \left< 12 \right>, K = \left< 18 \right>$, then $H \cap K$ would be the set of all integers divisible by $12$ and $18$. There's another name for that, which is for you to piece together.
b) Using the notation from part a), given two subgroups $H, K$ of $G$, we write $$H K : = \left< \left\{ h k : h \in H, k \in K \right\} \right> ,$$ i.e. $HK$ is the group generated by products of elements of $H, K$. Similar to a), when $G$ is abelian, we often write the group operation as addition, and say $$H + K : = \left< \left\{ h + k : h \in H, k \in K \right\} \right> .$$ In your situation, where $G = \mathbb{Z}, H = \left< 12\right>, K = \left< 18 \right>$, the question to ask is what is the set $$H + K = \left\{ 12 a + 18 b : a, b \in \mathbb{Z} \right\} ?$$ That's for you to figure out.