a) For any number $A$, let $\overleftarrow{A}$ be the reverse decimal expansion of $A$. Assume that $A = \overline{A_nA_{n-1}\cdots A_1}$ is a mirror-symmetry number with $m$ digits, all from $\{1, 2, 3\}$, and $A_n, A_{n-1}, \dots, A_1$ are blocks of $A$ with number of digits $m_n, \dots, m_1$ such that
$$A = \overleftarrow{A_n} \times \overleftarrow{A_{n-1}} \times \dots \times \overleftarrow{A_1}.$$
Note that for all $1 \le i \le n$, we have
$$\overleftarrow{A}_i \le \underbrace{333\cdots33}_{m_i} = \frac{10^{m_i} - 1}{3}.$$
On the other hand,
$$A \ge \underbrace{111\cdots11}_{m} = \frac{10^m - 1}{9}.$$
Therefore we obtain
$$\frac{10^m - 1}{9} \le \frac{10^{m_n} - 1}{3} \times \dots \times \frac{10^{m_1} - 1}{3}.$$
If $n \ge 2$ we have
$$\begin{aligned} 3^{n-2}(10^m - 1) &\le (10^{m_n} - 1)(10^{m_{n-1}} - 1)\cdots(10^{m_1} - 1) \\ &< 10^{m_n} \times 10^{m_{n-1}} \times \cdots \times 10^{m_2} \times (10^{m_1} - 1) \\ &< 10^{m_n+m_{n-1}+\cdots+m_1} - 1 = 10^m - 1. \end{aligned}$$
Which is impossible. Therefore $n=1$. So the only possible case is when $A = \overleftarrow{A}$, that means $A$ is a Palindromic number (a number that remains the same when its digits are reversed). Clearly, all Palindromic numbers with digits of $\{1, 2, 3\}$ satisfy the conditions.

b) This part is a test of effort! Note that if we could find a good number $m = \overline{A_1A_2\cdots A_n}$ where $A_i$'s are blocks of $m$ such that
$$\frac{m}{7} = A_1 \times \cdots \times A_n,$$
then $10m$ is also a good number because
$$\frac{10m}{7} = A_1 \times \cdots \times \overline{A_n 0}.$$
Therefore $m, 10m, 100m, \dots$ are all good numbers. So indeed, we just need to find a single good number. Now if we start to check the multiplies of $7$ one by one, we shall finally reach $7 \times 45 = 315$ that for which
$$\frac{315}{7} = 3 \times 15.$$
Therefore by putting $m = 315$, we can find infinitely many good numbers, $\{315, 3150, 31500, \dots\}$.

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