a) The largest number $m$ will have as many digits as possible, so we will choose the smallest possible digits. As next to a $1$ we can only put a $2$, and $2023 = 3 \cdot 674 + 1$, we choose $m = \underbrace{12121 \dots 121}_{1349 \text{ digits}}$, with $674$ digits equal to $2$ and $675$ digits equal to $1$.

b) As $8+9=17$ and $2022 = 17 \cdot 118+16$, if in the decimal writing of $n$ we would use at most $237$ digits, the sum of these digits would be at most $118 \cdot (8+9)+9 < 2022$, false. Therefore we can't use fewer than $2 \cdot 118 + 2 = 238$ digits. To use exactly $238$ digits, we must have $118$ groups of $8$ and $9$, and another two digits whose sum is $16$. The two digits could be $7$ and $9$, or $8$ and $8$. In both cases, the number of even digits of $n$ differs by $2$ from the number of its odd digits. This is impossible, because in the decimal writing of a special number, the even digits alternate with the odd ones, so the number of its even digits is either equal to that of its odd digits, or differs from it by $1$. Therefore $n$ must have at least $239$ digits. We are looking for three consecutive digits, smaller than $8$, whose sum is at least $16$ and which are the first three digits of $n$. Moreover, we are looking for the first digit of $n$ to be as small as possible. The smallest digits that fulfill these requirements are $5$, $6$, $7$. The solution is $n = 56787\underbrace{8989\dots89}_{117}$.