12
As long as the triangles are allowed to overlap, I think I can just Zorro through like this:
Counting:
Triangles with all 3 sides made of grid lines: 0
Triangles with 2 sides made of grid lines: 12 (4 per each drawn line)
Triangles with 1 side made of grid lines: 8 (4 for each intersecting red line pair)
Triangles with 0 sides made of grid lines: 0
As for ...
8
Unless I'm missing something obvious, the answer to the original question is:
And the answer to the bonus question is:
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