a) Consider the regular hexagon $ABCDEF$, centered at $O$, and let $T$, $U$, $V$, $X$, $Y$, $Z$ be the midpoints of the sides $AB$, $BC$, $CD$, $DE$, $EF$, $FA$, respectively.
Notice that the nodes situated on the sides of the hexagon cannot have trios, so a node that admits a trio is either $O$, or situated at distance $1$ from $O$. Denote these latter nodes by $M$, $N$, $P$, $Q$, $R$, $S$, as shown in Figure 1.
The side length of an equilateral triangle with nodes as vertices may be equal to $1$, $\sqrt{3}$, $2$, $\sqrt{7}$, $3$ or $2\sqrt{3}$. A sheet of side length $3$ or $2\sqrt{3}$ cannot be part of a trio, because the vertices of this sheet should be situated on the hexagon sides.
There are no sheets of side $\sqrt{7}$ that have $O$ as a vertex. Also, there are only two sheets of side $\sqrt{7}$ that have $R$ as a vertex, namely $RAU$ and $RCT$. Moreover, the intersection of a side $2$ sheet and a side $\sqrt{7}$ sheet of $R$ is non-empty, so we can't form trios with such sheets.
Consequently, the sheets RAU, RDP and REY (see Figure 1) form a trio with maximal possible area, and the maximal area is equal to
$$\frac{\sqrt{3}}{4} + \frac{3\sqrt{3}}{4} + \frac{7\sqrt{3}}{4} = \frac{11\sqrt{3}}{4}.$$

b) Considering the colored trio from Figure 2, we notice that OAB covers the sixth part of the hexagon. Rotating the view, OBC, OCD, ODE, OEF, and OFA are sheets for some other five trios of $O$. All these six trios cover the whole hexagon.
The node $R$ has exactly two sheets of maximal side length - which is equal to $\sqrt{7}$ - namely RAU and RCT. Since $RB = 3 > \sqrt{7}$, it follows that not all the points of the segment $RB$ can be covered with the trios of $R$.

c) We say that a trio is $(x, y, z)$-type if the side lengths of its sheets are equal to $x, y, z$.
The center $O$ has only $(1, \sqrt{3}, 2)$-type trios, as indicated in Figure 2. Now, we count how many trios have $OQR$ as one of its sheets. The only side $2$ sheets that we can choose to be part of a trio are $OAF$, $OAB$ and $OBC$. For each of these side $2$ sheets we can choose the side $\sqrt{3}$ sheet in two ways, so there are exactly $6$ trios which contain $OQR$. Since $O$ is the vertex of six side $1$ sheets, the node $O$ has exactly $36$ trios. The node $R$ admits only $(1, \sqrt{3}, 2)$ and $(1, \sqrt{3}, \sqrt{7})$ type trios.
Consider the sheets RDP and RNZ. To complete a trio, there are two choices for the side-$1$ sheet, specifically REX and REY. The same applies for the sheets RFM and RNV. Therefore, the node $R$ has exactly four $(1, \sqrt{3}, 2)$-type trios.
Likewise, the sheets RAU and RDP can form a trio together with one of the following three triangles of side length $1$: REX, REY and RSY. Also, there are three side $1$ sheets that form a trio with RCT and RFM. Consequently, $R$ has six $(1, \sqrt{3}, \sqrt{7})$-type trios.
It follows that there are $10$ trios associated to the node $R$, so there are $60$ trios associated to a node other than $O$. Adding all up, there are $96$ trios associated to the hexagon.