See fig for clarity.

 

radius of curvature = r = mv/qB

 

from the figure:

sin(a) = d/r = qBd/mv

Yes joyfrancis is right for q 39 just put in the values and you get the answers.

As for q 45,a.) the particles move in circular paths, as vel is perp to the magnetic field . t = 2 pi *m/ Bq

particles do not collide if their sum of radii <  d
The charges carried by them are same i.e q
hence d = 2 mv/qB
so v = qBd/2m
b.)v = vm/2
Min dist = r + r = 2r
so min dist = d/2
Max dist = d + 2r = 3d/2
c.)if v = 2vm

so vx = 2 vm cos x= 2vm cos(Bq/m)t

from symmetry collision takes place at d/2

so d/2 = integration of vx dt with limits from 0 to t = 2 vm integration of cos(Bqt/m) dt

hence d/2 = 2m vm/Bq sin (Bqt/m)

now we know tat vm = Bqd/2m

sin ( Bqt/m) = 1/2

Bqt/m = pi / 6

t = M pi/6Bq 
d.) luk up the answers its wonderfully explained there.
Hope its clear nudge me if u dont understand