There is only one monohex, as well as only one dihex and their total area is 3 hexes. If we combined them with the 3 trihexes, we obtained 12 hexes. Upon my request George Sicherman tested whether all the 310 dodecahexes, tilable with the hexiamonds, are also tilable with the combined sets of mono-, di- and trihexes. As it was expected, this answer is yes. This means that exactly 310 shapes are bi-tilable with the hexiamonds and the mono-, di- and trihex sets. Here are some examples (not necessarily unique solutions shown):
A stronger version of the same problem would be to require the monohex to be exactly in the same place as the hexagon hexiamond. This is probably also always possible, considering that the hexagon hexiamond is seldom off the imaginary hexagonal grid laid by the dodecahex structure: