All right, already. Never let it be said I copped out of a math problem.
(1) Sketch the quadrilateral ABCD with the information given. It’s just two 30-60-90 triangles joined symmetrically along the hypoteneuse. All the angles are 30-60-90-120, all the lengths combinations of 1, 2, and sqr(3).
(2) Question (i) is now trivial, from symmetry. (Though of course you’d have to find a politer way to say that in an examination paper.)
(3) For question (ii), again from symmetry, it is sufficient to calculate the angle A1-E-C1, which can easily be done from the lengths involved, since this is yet another 30-60-90 triangle. Answer: 90 degrees.
(4) For question (iii), first parallel-transport the triangle B-C-C1 until B coincides with A. You now have a solid angle at A formed by three intersecting planes, two of them–both with angle 30 degrees at the apex–at right angles to each other (i.e. the planes A-B-C-D and B-C-C1).
(One of those 30 angles is C-B-C1. The other you get from parallel-transporting BC until B coincides with A, since angle DAB is 120 and angle ABC is 90.)
Chopping off these two planes by a vertical one to make 30-60-90 triangles with sides length 1-2-sqr(3), one other face of the tetrahedron you just made–the one with an angle at the vertex A–has sides 2, 2, sqr(2). Its vertex angle is therefore 90 degrees.