Solution:

a.
Let the set $A$ consist of the $4 n$ integers $1, 2, \ldots, 4 n$ and let the set $B$ consist of the $2 n$ even integers $4 n + 2, 4 n + 4, \ldots, 8 n$. We claim that the $6 n$-element set $S = A \cup B$ has the desired property.

Indeed, the least common multiple of two (even) elements of $B$ is no larger than $8 n \cdot (8 n / 2) = 32 n^{2}$, and the least common multiple of some element of $A$ and some element of $A \cup B$ is at most their product, which is at most $4 n \cdot 8 n = 32 n^{2}$.

b.
We prove the following lemma: "If a set $U$ contains $m + 1$ integers, where $m \geqslant 2$, that are all not less than $m$, then some two of its elements have least common multiple strictly larger than $m^{2}$."

Let the elements of $U$ be $u_{1} > u_{2} > \cdots > u_{m+1} \geqslant m$. Note that $1 / u_{1} \leqslant 1 / u_{i} \leqslant 1 / m$ for $1 \leqslant i \leqslant m+1$. We partition the interval $[1 / u_{1} ; 1 / m]$ into $m$ subintervals of equal length. By the pigeonhole principle, there exist indices $i, j$ with $1 \leqslant i < j \leqslant m+1$ such that $1 / u_{i}$ and $1 / u_{j}$ belong to the same subinterval. Hence
$$0 < \frac{1}{u_{j}} - \frac{1}{u_{i}} \leqslant \frac{1}{m} \left( \frac{1}{m} - \frac{1}{u_{1}} \right) < \frac{1}{m^{2}}.$$
Now $1 / u_{j} - 1 / u_{i}$ is a positive fraction with denominator $\operatorname{lcm}(u_{i}, u_{j})$. The above thus yields the lower bound $\operatorname{lcm}(u_{i}, u_{j}) > m^{2}$, completing the proof of the lemma.

Applying the lemma with $m = 3 n$ to the $3 n + 1$ largest elements of $T$, which are all not less than $3 n$, we arrive at the desired statement.

A Variant.

Alternatively, for part (b), we prove the following lemma: "If a set $U$ contains $m \geqslant 2$ integers that all are greater than $m$, then some two of its elements have least common multiple strictly larger than $m^{2}$."

Let $u_{1} > u_{2} > \cdots > u_{m}$ be the elements of $U$. Since $u_{m} > m = m^{2} / m$, there exists a smallest index $k$ such that $u_{k} > m^{2} / k$. If $k = 1$, then $u_{1} > m^{2}$, and the least common multiple of $u_{1}$ and $u_{2}$ is strictly larger than $m^{2}$. So let us suppose $k > 1$ from now on, so that we have $u_{k} > m^{2} / k$ and $u_{k-1} \leqslant m^{2} / (k-1)$. The greatest common divisor $d$ of $u_{k-1}$ and $u_{k}$ satisfies
$$d \leqslant u_{k-1} - u_{k} < \frac{m^{2}}{k-1} - \frac{m^{2}}{k} = \frac{m^{2}}{(k-1)k}$$
This implies $m^{2} / (d k) > k-1$ and $u_{k} / d > k-1$, and hence $u_{k} / d \geqslant k$. But then the least common multiple of $u_{k-1}$ and $u_{k}$ equals
$$\frac{u_{k-1} u_{k}}{d} \geqslant u_{k} \cdot \frac{u_{k}}{d} > \frac{m^{2}}{k} \cdot k = m^{2}$$
and the proof of the lemma is complete.

If we remove the $3 n$ smallest elements from set $T$ and apply the lemma with $m = 3 n$ to the remaining elements, we arrive at the desired statement.