a.
See Problem B.4.

b.
Let it be possible to choose six four-digit numbers satisfying the problem condition. By the above, any of digits $1$, $2$, $3$, $4$, $5$, $6$, $7$, $8$ must be exactly in three of these six numbers.

Further, it is not hard to prove that no two of the chosen numbers have the same three digits.

Let $a_1, a_2, a_3, a_4, a_5, a_6$ be the chosen six numbers. Let $P_i$ denote the set consisting of six pairs formed by the digits of $a_i$, $i = 1, \dots, 6$. It is easy to see that the set $\bigcup_{i=1}^6 P_i$ has exactly $28$ elements. Moreover, either $P_i \cap P_k = \emptyset$ or $P_i \cap P_k$ has exactly one element, and $P_i \cap P_j \cap P_k = \emptyset$, for all distinct $i, k, j$.

By inclusion-exclusion formula ($N(X)$ is the number of elements of the set $X$)
$$N\left(\bigcup_{i=1}^{6} P_i\right) = \sum_{i=1}^{6} N(P_i) - \sum_{\substack{i,j=1 \\ i<j}}^{6} N(P_i \cap P_j). \quad (*)$$
Since $N\left(\bigcup_{i=1}^{6} P_i\right) = 28$ and $N(P_i) = 6$ for any $i = 1, \dots, 6$, equality $(*)$ can be written as
$$\sum_{\substack{i,j=1 \\ i < j}}^{6} N(P_i \cap P_j) = 8. \qquad (**)$$
Let $x$ be the number of pairs $(P_i, P_j)$, $i < j$, for which $N(P_i \cap P_j) = 0$, then for the remaining $15-x$ pairs we have $N(P_i \cap P_j) = 1$. Thus $(*)$ can be rewritten as $15-x=8$, or $x=7$.

It is not very hard to prove that $x \neq 7$. Thus $x \leq 6$, and so $N \geq 7$.