(a) We first show that there must be a non-acute triangle among any 4 points. Consider the convex hull of $A$, $B$, $C$, $D$.

* If the convex hull is a quadrilateral $ABCD$, then since
$$\angle ABC + \angle BCD + \angle CDA + \angle DAB = 360^\circ,$$
one of these angles is at least $90^\circ$. This gives rise to a non-acute triangle.

* If the convex hull is a triangle, say $\triangle ABC$, then since
$$\angle ADB + \angle BDC + \angle CDA = 360^\circ,$$
one of these angles is obtuse.

Now, suppose on the contrary that at most 2 triangles are non-acute. Note that each triangle belongs to exactly 2 quadrilaterals, and there are $\binom{5}{4} = 5$ quadrilaterals in total. Therefore, there must be a quadrilateral which does not consist of any non-acute triangles, contradicting the above observation. Therefore, there are at least 3 non-acute triangles.

(b) There are $\binom{100}{5}$ groups of 5 points formed from the 100 points. By part (a), there are at least 3 non-acute triangles in each group. Since each triangle belongs to $\binom{97}{2}$ groups of 5 points, there are at least
$$3 \binom{100}{5} \div \binom{97}{2}$$
non-acute triangles. As there are $\binom{100}{3}$ triangles in total, at least
$$3 \binom{100}{5} \div \binom{97}{2} \div \binom{100}{3} = \frac{3}{10} = 30\%$$
of the triangles are non-acute.