8. (a) Consider $$a_{i}=i k+1, \quad i=1,2, \ldots, m ; \quad b_{j}=j m+1, \quad j=1,2, \ldots, k$$ Assume that $m k \mid a_{i} b_{j}-a_{s} b_{t}=(i k+1)(j m+1)-(s k+1)(t m+1)=$ $k m(i j-s t)+m(j-t)+k(i-s)$. Since $m$ divides this sum, we get that $m \mid k(i-s)$, or, together with $\operatorname{gcd}(k, m)=1$, that $i=s$. Similarly $j=t$, which proves part (a). (b) Suppose the opposite, i.e., that all the residues are distinct. Then the residue 0 must also occur, say at $a_{1} b_{1}: m k \mid a_{1} b_{1}$; so, for some $a^{\prime}$ and $b^{\prime}, a^{\prime}\left|a_{1}, b^{\prime}\right| b_{1}$, and $a^{\prime} b^{\prime}=m k$. Assuming that for some $i, s \neq i$, $a^{\prime} \mid a_{i}-a_{s}$, we obtain $m k=a^{\prime} b^{\prime} \mid a_{i} b_{1}-a_{s} b_{1}$, a contradiction. This shows that $a^{\prime} \geq m$ and similarly $b^{\prime} \geq k$, and thus from $a^{\prime} b^{\prime}=m k$ we have $a^{\prime}=m, b^{\prime}=k$. We also get (1): all $a_{i}$ 's give distinct residues modulo $m=a^{\prime}$, and all $b_{j}$ 's give distinct residues modulo $k=b^{\prime}$. Now let $p$ be a common prime divisor of $m$ and $k$. By $(*)$, exactly $\frac{p-1}{p} m$ of $a_{i}$ 's and exactly $\frac{p-1}{p} k$ of $b_{j}$ 's are not divisible by $p$. Therefore there are precisely $\frac{(p-1)^{2}}{p^{2}} m k$ products $a_{i} b_{j}$ that are not divisible by $p$, although from the assumption that they all give distinct residues it follows that the number of such products is $\frac{p-1}{p} m k \neq \frac{(p-1)^{2}}{p^{2}} m k$. We have arrived at a contradiction, thus proving (b).