period of the mass?

1) See picture at attachment.
Mass 2 is stationary, so ΣF of mass 2=0. So, T1=W <=> T1=3*g
T2=T1 because T2 is the reaction of T1
T3=T2 (otherwise the rope would break)
T4=T3=3*g because T4 is the reaction of T3.

T4 is the centripetal force.
So, the centripetal acceleration should be: T4/m2=(3/2)*g and if we assume that g=9,8:
ac=14,7m/s2

And using the formula, we have:
T=square root of: (4π2*R/ac)=square root of: [4,935/14,7]=0,58 seconds

2)
We have: the acceleration of mass m is the same as the acceleration of the mass 3m (the rope is inextensible and not loose)
In addition, T1=T4 (for the same reason as the previous exercise)
The acceleration of mass m is: (W-T1)/m
The acceleration of  mass 3m is: T4/3m
So, we have (W-T1)/m=T4/3m <=> 3*(m*g-T1)=T4 <=> 3*m*g-3T1=T1 <=> 4T1=3*m*g <=> T1=(3/4)*m*g.

We already said that a=T4/3m (or a=(W-T1)/m , you can use whichever equation you want)
So: a=T4/3m <=> a=T1/3m <=> a=[(3/4)*m*g]/3m <=> a=(1/4)*g or a=2,45m/s2 if we assume that g=9,8m/s2.

(See attachment for pictures for both exercises)