Solution:

(a) Let us fix $n$ consecutive numbers $a_{1}, a_{2}, \ldots, a_{n}$. Now let $d$ be a number larger than $n$ that has no common factors with any of $a_{1}, a_{2}, \ldots, a_{n}$ (for instance, a prime number larger than $a_{n}$), and set $b_{1} = a_{1} + d, b_{2} = a_{2} + d, \ldots, b_{n} = a_{n} + d$; then $a_{1}, a_{2}, \ldots, a_{n}$ and $b_{1}, b_{2}, \ldots, b_{n}$ are two sequences of length $n$ that are disjoint from each other and such that, for every $i$ between 1 and $n$, the greatest common divisor of $a_{i}$ and $b_{i}$ is equal to 1: indeed, if a common factor of $a_{i}$ and $b_{i}$ existed, it would also be a factor of $b_{i} - a_{i} = d$, and this is impossible, because $d$ and $a_{i}$ have no common factors.

(b) Similarly as before, let us fix $a_{1}, a_{2}, \ldots, a_{n}$ consecutive, with $a_{1} > 1$; let us then take an integer $d > n$ that has factors in common with each of the elements $a_{1}, \ldots, a_{n}$ (for instance, the product $a_{1} a_{2} \cdots a_{n}$), and set $b_{1} = a_{1} + d, b_{2} = a_{2} + d, \ldots, b_{n} = a_{n} + d$. Once again the sequences $a_{1}, a_{2}, \ldots, a_{n}$ and $b_{1}, b_{2}, \ldots, b_{n}$ are disjoint from each other, but in this case $a_{i}$ and $b_{i}$ always have a common factor: indeed $a_{i}$ and $d$ always have a common factor, and hence the same is true for $a_{i}$ and $d + a_{i} = b_{i}$.