In the first case, the total weight
$$1 + 2 + \cdots + 2010 + 2011 = \frac{2011 \cdot 2012}{2}$$
is not a multiple of $3$. Therefore, there is no solution in this case.

In the second case, we distribute the first $8$ parcels as follows: Parcels $1$, $2$, $3$, $6$ (of total weight $12$) are put into the first group. Parcels $4$ and $8$ (also of total weight $12$) are put into the second group. Parcels $5$ and $7$ (of total weight $12$) are put into the third group.

The remaining $2004$ parcels $9$, $\ldots$, $2012$ are first divided into $334$ blocks of consecutive integers $6k + 3$, $6k + 4$, $6k + 5$, $6k + 6$, $6k + 7$, $6k + 8$ for $1 \le k \le 334$. Of each block, $6k + 3$ and $6k + 8$ (of weight $12k + 11$) are put into the first group. Parcels $6k + 4$ and $6k + 7$ are put into the second group. Finally, parcels $6k + 5$ and $6k + 6$ are put into the third group. This yields a valid partition. $\square$