You’re right, the intersection can be different from the product. Just consider the left ideals of $R=M_2(F)$ given by $Re_{11}$ and $Re_{22}$. So that does not hold at all, even in a simple ring.
But surely you’ve covered the proposition that an Artinian prime ring is simple? From that, it is easy, because for any prime ideal $P$ in a left Artinian ring, $R/P$ is simple, hence $P$ is maximal.
No, ideal product doesn't usually distribute over intersection. You'll need something like the Artin-Rees lemma (see [Stacks 00IN]) to prove that a local Noetherian ring $(R, \mathfrak{m})$ is $\mathfrak{m}$-adically separated.
In your notation, when $m$ is a flat ideal and $A_\alpha$ is a finite set of ideals, then multiplication by $m$ does distribute over the intersection $\bigcap A_\alpha$ (more generally its true if the family $A_\alpha$ is in an appropriate sense compact). For a reference, see e.g. [Stacks 0BBY].
When we don't require that $m$ is flat, this can really fail, even when $m$ is the maximal ideal of a local Noetherian ring.
As an example, consider the ring $k[x,y]$ with $k$ a field and the maximal ideal $\mathfrak{m} = (x,y)$. Let $R = k[x,y]_\mathfrak{m}$. This is a local Noetherian ring with maximal ideal $(x,y)$. The ideals $m = (x,y)$, $A = (x), B = (y)$ satisfy the following identies:
$$mA \cap mB = (xy)$$
$$m(A \cap B) = xy(x,y)$$
As a closing remark, let's note the maximal ideal being flat is a big deal. Observe that in a Noetherian local ring a flat ideal is automatically free (Noetherian implies f.g. flat, and local implies f.g. flat is free), and in general an ideal is free iff it is principally generated by a non-zero divisor.
So in a Noetherian local ring, the maximal ideal is flat iff it is principally generated by a non-zero divisor. This is a lot to ask for. It implies that every ideal is a power of $m$. Hence the ring is either a discrete valuation ring or a field.
Best Answer
To see $IJ\supseteq I\cap J$ you should use the assumption on $I$ and $J$. Take $x\in I$ and $y\in J$ so that $x+y=1$. Now, suppose $a\in I\cap J$, then $ax+ay=a$. $ax\in IJ$ because $a\in J$ and $x\in I$ and similarly $ay\in IJ$ because $a\in I$ and $x\in J$. So, $ax+ay\in IJ$ so, $a\in IJ$.