Example of $H$ and $K$ and find the order of $HK$ when $o(H \cap K) >1$. Can you give an example of two subgroups $H$ and $K$ where $o(H \cap K) > 1$ and hence find the order of $HK$?
 A: Your question is ambiguous: 
$(1)$ Are $H$ and $K$ both subgroups of the same group $G$? If so, Are you given any information about $G$?  
$(2)$ And what do you mean by $HK$? Assuming you mean $HK$ is a subgroup of the same group of which $H$ and $K$ are subgroups, then perhaps you mean for $HK = \{h*k | h\in H \;\text{and}\;k\in K\}$?
$(3)$ Also "O(H and K)" is ambiguous. Do you mean that the $\text{ord}(H) > 1$ and $\text{ord}(K) > 1$? Or do you mean $\text{ord}(H \cap K) > 1$?
CLARIFIED "O(H and K) = $\text{ord}(H \cap K)$...

That said:
First, assuming $H \le G$ and $K \le G$ ($\le$ meaning "is a subgroup of"), then $H \cap K \le G$. But that intersection may be only the identity of $G$, and hence of order 1. $|H \cap K| = \text{ord}(H\cap K) > 1$ if and only if $\text{gcd}(|H|, |K|) \neq 1$. 
We can show that if $H$ and $K$ are subgroups of an abelian group $(G, *)$, and if $HK$ denotes a subgroup, then
$$HK = J = \{h*k | h\in H \;\text{and}\;k\in K\} \le G.$$ So if the order of $H = a >1$, and the order of $K = b > 1$, then the order of  $HK = J = \text{lcm}(a, b)$. 
Can you think of a finite abelian group $G$ with subgroups $H, K$ such that if you know $|H| = a > 1$ and $|K| = b > 1$, then $|HK|>1$?
Take, for example, $G = \mathbb{Z}_{24}, \;H = \langle 4\rangle = \{0, 4, 8, 12, 16, 20\} \; K = \langle 6 \rangle = \{0, 6, 12, 18\},\; H\cap K = \{0, 12\}$.
Clearly, $H \le G, \; K \le G$. $|H| = 6>1, \; |K| = 4>1, \; |H\cap K| = 2 > 1$. 
Then $HK = \{h+k|h \in H, \; k \in K\} = \langle 10 \rangle = \langle 2 \rangle = \{0, 2, 4, 6, 8, 10, 12, 14, 16, 18, 20, 22\}$ and $|\langle 2 \rangle | = 12 = \text{lcm(6, 4)} > 1$.

Note: you can also take for an example any $H, K \le \mathbb{Z}$. 
