Look closer at the specs of your relay board. I believe that you have 2 power sources. The first one is the VCC for the relay side of the optic isolators. For that you can use 5VDC. The other VCC (on the row of input pins) would be the 3.3V VCC that the imp uses. The opto isolators are the separation. I also believe that the opto isolators are designed such that they float high (the relays are off). If you sink the pins to ground, the relay turns on. This logic might seem backwards from what you would think: 0=relay on, 1=relay off
The imp pins can sink current much more than source. So I would not think it’s a problem. In my opinion, it’s always better to sink a floating high, than to source. Larger loads might require a basic 2N222 NPN switching transistor.
The whole point to this board is the optic isolators. They control the relays with a larger current 5VDC VCC and isolate that from the lower current digital input pins.
But, let Hugo answer too … I would hate to have given a wrong answer and things go bad.
… I have burned-up an Imp before, it’s not fun … [sigh]