In fact, the inclusion $IJ \subseteq I \cap J$ is always true. Given ideals $I$ and $J$, an element of $IJ$ is of the form $x = a_1 b_1 + \cdots + a_t b_t$ for some $t \in \mathbb{Z}_{\geq 0}$ and $a_k \in I$ and $b_k \in J$. Since ideals are closed under arbitrary multiplication, then each term of the above sum is in both $I$ and $J$, and since ideals are closed under addition, then the sum is in $I$ and $J$, so $x \in I \cap J$.
Now, let's consider the reverse inclusion. For ease of notation, let's just consider two primary ideals $Q_1$ and $Q_2$. By part (b) we have $Q_1 = (p_1^m)$ and $Q_2 = (p_2^n)$ for some primes $p_1, p_2$ and, as per my comment, assume $p_1$ and $p_2$ are non-associate. Given $x \in Q_1 \cap Q_2$, then $x = p_1^m a$ and $x = p_2^n b$ for some $a,b \in R$. Then $p_1 \mid p_2^nb$, so $p_1 \mid p_2^n$ or $p_1 \mid b$. If $p_1 \mid p_2^n$, then by repeated using the fact that $p_1$ is prime we find that $p_1 \mid p_2$. Then $p_2 = p_1 c$ for some $c \in R$. But since primes are irreducible, then $c$ must be a unit, which contradicts that $p_1$ and $p_2$ are non-associate. Thus $p_1 \mid b$, so $b = p_1 b_1$ for some $b_1 \in R$.
Now we have
$$
p_1^m a = x = p_2^n b = p_2^n p_1 b_1 \, .
$$
Since $R$ is a domain, then we cancel the factor of $p_1$, which yields $p_1^{m-1} a = p_2^n b_1$. Repeating the same argument as above (or by induction, if you want to be rigorous), then $p_1 \mid b_1$ so $p_1^2 \mid b$, and so on and so forth until we get $p_1^m \mid b$. Then $b = p_1^m b_m$ for some $b_m \in R$, so
$$
x = p_2^n b = p_2^n p_1^m b_m \in (p_1^m)(p_2^n) = Q_1 Q_2 \, .
$$
We can use the above in the induction step to prove this result for any finite number of primary ideals. I'll leave the details to you.
Suppose we know the result for $k$ primary ideals. Then \begin{align*} Q_1 \cap \cdots \cap Q_k \cap Q_{k+1} &= (Q_1 \cap \cdots \cap Q_k) \cap Q_{k+1} = (Q_1 \cdots Q_k) \cap Q_{k+1} \, . \end{align*}
Given $x \in \bigcap_{i=1}^{k+1} Q_i = (Q_1 \cdots Q_k) \cap Q_{k+1}$, then $$x = p_1^{n_1} \cdots p_k^{n_k} a \qquad \text{and } \qquad x = p_{k+1}^{n_{k+1}} b $$ for some $a,b \in R$. You can use the same proof as above to show that $p_{k+1}^{n_{k+1}} \mid a$ which will finish the induction.
The fact that prime ideals are maximal in an Artinian ring is a very frequently asked duplicate.
Assuming all the $p_i$ are prime (as the notation suggests) and hence maximal, they satisfy your setup. Think of what it would mean for $a_n\subseteq p_j$ for some $j>n$. It would mean that $\prod_{i\leq n} p_i\subseteq a_n\subseteq p_j$, in which case $p_i\subseteq p_j$ for some $i\leq n$. This has been ruled out by hypothesis though.
So in fact, $a_n$ is not contained in any such $p_j$, therefore $a_n\cap p_j$ is strictly contained in $a_n$. This amounts to saying that if there were infinitely many such $p_i$, they would necessarily produce a strictly descending chain with members $a_n$. Since that is not possible, the family of primes can only be finite.
(There are certainly duplicates for the question itself but I could not find any using the strategy requested. They could still be out there.)
b) as you said is obvious given a)
c) Assuming $m_A$ is the jacobson radical, equal to the intersection of all prime/maximal ideals in this scenario, you have immediately by the Chinese Remainder theorem that $A/m_A\cong \prod_P A/P$ where $P$ ranges over the prime ideals. Since each prime $P$ is maximal, these are all fields, and we've established that there are only finitely many.
Best Answer
If $P_i \subset P_j$ for every $j$, then $P_1 \cap \cdots \cap P_n = P_i$ is prime. Conversely, if $P_1 \cap \cdots \cap P_n = Q$ is prime then $Q \subset P_j$ for every $j$ and also there exists some $i$ such that $P_i \subset Q$. Hence $Q=P_i$. Here i used the fact that if $a$ is an ideal such that $a \supset p_1 p_2 \cdots p_m$ where the $p_i$ are prime, then $a$ contains some $p_k$. (If i am not mistaken, this is proposition 1.11b in Atiyah-MacDonald.)