In an acute triangle, all angles are less than 90 degrees.
Source: Martin Gardner's Mathematical Diversions, Used in Ken's Puzzle of the Day 4/8/94.
1) It is possible to divide any triangle in acute triangles with the
following method:
Let consider the triangle ABC. If ABC is acute,the problem is obviously
solved. If not (for example angle(BAC) > 90o or = 90o), let D on AB, E and F
on BC, G on AC in order to have 2 acute triangles BDE and CFG and a pentagon
ADEFG. If this pentagon is close to a "regular" pentagon, a point O inside
this pentagon will determine 5 acute triangles OAD, ODE, OEF, OFG and OGA. We
infer that at least 7 acute triangles are necessary to divide a triangle with
a obtuse or right angle.
2) To divide a square in acute triangles, it is easy to divide the square in 2 right isoceles triangles and we are taken back to the former question.
For part #2, I think there is a well known solution requiring only 8 triangles. If the square is ABCD, bisect BC at E, and bisect DA at F. Draw semicircles inside the square on AB and BE, insersecting at G. Pick any point inside the region made by line EF and arcs EG and GF; call this G'. Since G' is outside of both semi-circles, the angles BG'E and BG'A are both acute. Let G'' be the reflection of G' across EF. Then CG''E and CG''B are also acute.
The rest is trivial to show that BG'E, G'G''E, EG''C, G''BF, G'G''F, AG'F, and AG'B are all acute.