Exercise 5.32 (a) Note that the Edgeworth box is square since e1+e2=(10,10). Also the utility functions are rectangular hyperbolas and these are symmetric around the straight line running across the box from the bottom left to the top right. This line is the contract curve.
(b)and(c) We have U1q(8,2)=U2q(2,8)=16. The core is 4<=x1<=6 and 4<=x2<=6 with x1=x2. You can read this off the diagram.
(d) Form the coalition {11,12,21} with endowment (18,12). Now split this to give both consumers of type 1 an allocation of (6,3) and the consumer of type 2 an allocation of (6,6). Check this is feasible for this coalition and that it is strictly preferred to x11=x12=(4,4) for consumers of type 1 and (obviously) indifferent to (6,6)for the consumer of type 2. Tweak it a little to say (6,2.9)for the two type 1's and (6,6.2) for the one type 2 to get strict preference for all three consumers in the coalition.
This is a standard coalition and allocation for this type of problem even if the problem does not have the nice symmetry of this example. The coalition is two consumers of one type and one of the other (Not much else is possible, if you think about it). The allocation is obtained by taking the initial endowment and averaging it with the point on the contract curve indifferent to the initial endowment. In this case it is the average of (8,2)and (4,4)to get(6,3). However, the resultant allocation for all 4 consumers is not in the core (See Theorem 5.16)
Hope this helps.