For your first question. Suppose $(A,d,m,m_3,m_4\dots)$ is an $A_\infty$-algebra. The operation $m_3$ gives a homotopy between $m(-,m(-,-))$ and $m(m(-,-),-)$, which I will abusively denote as $a(bc)$ and $(ab)c$.

Now consider the two operations $A^{\otimes 4} \to A$ given by $a(b(cd))$ and $((ab)c)d$. Using $m_3$, you have two ways of going from the first to the second:

- Use $m_3$ three times to create a homotopy:
$$a(b(cd)) \to a((bc)d) \to (a(bc))d \to ((ab)c)d.$$
- Use $m_3$ twice to create a homotopy:
$$a(b(cd)) \to (ab)(cd) \to ((ab)c)d.$$

If you combine these two homotopies, you get two classes of degree $2$ maps $A^{\otimes 4} \to A$. These two maps have no reason to be homotopic. Well, the higher operation $m_4$ gives a homotopy between these two homotopies!

The even higher operations $m_5$, $m_6$ and so on work the same way. This is very nicely encoded in Stasheff's associahedra. The first two associahedra is just a point, representing the identity and $m_2$; the next one is a segment, representing $m_3$, a homotopy between $a(bc)$ and $(ab)c$; the next one is a pentagon, whose edges are the five arrows I drew above; and so on.

For your second question, the answer is Massey products. Very briefly, suppose that you have a differential graded algebra $A$ and three cycles $a,b,c$ such that $ab = d\alpha$ and $bc = d\beta$. Then the class $abc$ vanishes "in two different ways", because $abc = d(\alpha c) = d(a \beta)$. It follows that $\alpha c - a \beta$ is a homology class, called the triple Massey product $\langle a,b,c \rangle$. The operation $m_3$ can be used to represent this triple Massey product on homology. It's a bit technical to explain how, and the explanation involves the Homotopy Transfer Theorem.

For both answers, I think a good reference that cites pretty much all the other possible ones is the book *Algebraic Operads* by Loday and Vallette.