As Henry and Vercassivelaunos point out in the comments, the trick to answering your second question is to consider $\mathbb Q$, which you can show is not complete but is dense. (Terminology: A linear order $<$ is said to be dense if whenever $a<b$ there is some $z$ with $a<z<b$.)
My hint for your first question is to also try to prove that the answer is no. In this case, a counterexample would be a linear order $(A,<)$ which satisfies $\forall a,b\in A: a\le b\implies\exists c\in A:a\le c\le b$ but isn't dense. My advise is to start with your favorite non-dense linear order, and check to see if it satisfies this second property.
Edit: Getting back to your second question, not only are there dense linear orders which are not complete (like $\mathbb Q$), there are also complete linear orders which are not dense. Unlike in the previous paragraph I can't promise that your favorite non-dense linear order is a counterexample (maybe you have gross favorites), but I think you can show e.g. $\mathbb N$ is complete (by your definition) but not dense.
Edit 2: Above I took your questions as asking if the implications held over linear orders, since the question doesn't talk about fields. But it's also possible that you meant to ask in the context of ordered fields (which I think is the usual context when talking about the completeness axiom). In that case the answer to your second question is still no because of $\mathbb Q$, but the answer to your first question is yes. In fact:
Exercise 1: Show that every ordered field has the property that $\forall a,b\in A: a\le b\implies\exists c\in A:a\le c\le b$.
Exercise 2: Show that every ordered field is dense.
The first exercise should be easy if you've already checked that your favorite non-dense linear order satisfies said property. My hint for the second is, for $a<b$, to show that $z=\frac{a+b}{1+1}$ is well-defined (which is not obvious from just the field axioms!) and satisfies $a<z<b$.
