a) in D₄h
 
 

First, label axes and location of symmetry elements:

 

D4h	E	2C4	C2	2C2'	2C2''	i	2S4	h	2v	2d
a1g	1	1	1	1	1	1	1	1	1	1
a2g	1	1	1	–1	–1	1	1	1	–1	–1
b1g	1	–1	1	1	–1	1	–1	1	1	–1
b2g	1	–1	1	–1	1	1	–1	1	–1	1
eg	2	0	–2	0	0	2	0	–2	0	0
a1u	1	1	1	1	1	–1	–1	–1	–1	–1
a2u	1	1	1	–1	–1	–1	–1	–1	1	1
b1u	1	–1	1	1	–1	–1	1	–1	–1	1
b2u	1	–1	1	–1	1	–1	1	–1	1	–1
eu	2	0	–2	0	0	–2	0	2	0	0

Red. Rep.

1

–1

1

–1

1

1

–1

1

–1

1

The reducible representation exactly matches the b_2g irreducible representation, so no further work is required.
(b_1g is also possible, depending upon how the x and y axes are located.)  Also notice that the object looks just like
a d orbital, in particular the d_xy orbital, which transforms as b_2g!
 
 

b) in D_6h

First, label axes and location of symmetry elements:

 

D6h	E	2C6	2C3	C2	3C2'	3C2''	i	2S3	2S6	h	3d	3v
a1g	1	1	1	1	1	1	1	1	1	1	1	1
a2g	1	1	1	1	–1	–1	1	1	1	1	–1	–1
b1g	1	–1	1	–1	1	–1	1	–1	1	–1	1	–1
b2g	1	–1	1	–1	–1	1	1	–1	1	–1	–1	1
e1g	2	1	–1	–2	0	0	2	1	–1	–2	0	0
e2g	2	–1	–1	2	0	0	2	–1	–1	2	0	0
a1u	1	1	1	1	1	1	–1	–1	–1	–1	–1	–1
a2u	1	1	1	1	–1	–1	–1	–1	–1	–1	1	1
b1u	1	–1	1	–1	1	–1	–1	1	–1	1	–1	1
b2u	1	–1	1	–1	–1	1	–1	1	–1	1	1	–1
e1u	2	1	–1	–2	0	0	–2	–1	1	2	0	0
e2u	2	–1	–1	2	0	0	–2	1	1	–2	0	0

Red. Rep.

2

1

–1

–2

0

0

–2

–1

1

2

0

0

The reducible representation exactly matches the e_1u irreducible representation, so no further work is required.
(The characters for C₆ and C₃ are found as cos60^o and cos120^o, respectively.)  In this case the two objects look
a lot like modestly distorted p orbitals, in particular p_x and p_y, which transform as a pair in the e_1u representation.
 

c) All of the Cartesian coordinates for all of the atoms in NH₃.
 
 

The structure of ammonia is pyramidal, which belongs to the C_3v point group. Labeling the axes (use a similar labeling scheme for all atoms) gives:

 

C3v	E	2C3	3v
a1	1	1	1
a2	1	1	–1
e	2	–1	0

 
 

Red. Rep.

12

0

2

 

h = 6

n(a₁) = (1/6)[(1)(1)(12) + (2)(1)(0) + (3)(1)(2)] = 3

n(a₂) = (1/6)[(1)(1)(12) + (2)(1)(0) + (3)(–1)(2)] = 1

n(e) = (1/6)[(1)(2)(12) + (2)(–1)(0) + (3)(0)(2)] = 4

3a₁ + a₂ + 4e (this checks with the total dimensionality: 3 + 1 + 4(2) = 12)