Annie,
the first one:
B-BOY G-GIRL (g1g2g3g4)-a group of 4 girls supposed to sit together
First we consider 4 girls supposed to sit together as a group represented by (g1g2g3g4). Apart from said 4 girls, there are 4 boys waiting outside. We can simply consider those 4 plus the only group of girls; all 8 students can be arranged like, for instance, [b1 (g1g2g3g4) B2 B3 B 4], of which B1, (g1g2g3g4), B2, B3, B 4 are regarded as 5 distinctive elements. The permutation for all 5 elements is A(5,5); permutation in the group (g1g2g3g4) is A(4,4). Thus the product A(5,5)*A(4,4) implies the overall permutation.
The total permutation for all 8 people to be arranged randomly is A(8,8), and thus we can simply get the ans:
A(5,5)*A(4,4)/A(8,8)=1/14
the second one:
let MA/AB=k, similarly AN/AC=k, and we get MA=kAB, AN=kAC
using S=1/2 ab sin A, we get
1/2 AM*AN*sin A=s (1)
1/2 AB*AC*sin A=2s (2)
let (1)/(2)
then [k(AB)*K(AC)]/(AB*AC)=1/2
hence k^2=1/2, k=1/sqrt(2)
silentgrass.