For 2ai)
g=F/m
Therefore: F=gm
F=(mv^2)/r
Equating these gives gm=(mv^2)/r
So v^2=gr as the two m's cancel. This gives v=7445ms^-1, which is 7400ms^-1 to 2sigfig (unlike the other guy's answer)
For 3bii) g=-GM/r^2
-GM is constant, so g is proportional to 1/r^2
Therefore g[at M] / g[at A]=(1/r[at M]^2) / (1/r[at A]^2)
So to find gravitational field strength at M, g[at M]:
g[at M] = g[at A] * ((1/(orbital radius at M)^2) / (1/(orbital radius at A)^2))
= 7.5 * ((1/(2.4*10^10)^2)/(1/(1.3*10^8)^2))
= 2.22*10^-4 Nkg^-1