Even review problem solutions from Buchmann:

8.7.4:
>>> factoring.factor(899) # find p, q
[29, 31]
>>> 28*30 # (p-1)*(q-1)
840
>>> from xgcd import xgcd
>>> xgcd(11, 840) # find inverse of 11 mod (p-1)(q-1)
(1, -229, 3)
>>> -229 % 840  # positve representative for secret key
611
>>> pow(468, 611, 899) # decrypt
13
>>> pow(13, 11, 899) # check - OK
468

so the answer is 13

8.7.10:  A bit of a challenge since the algorithm is not given to you.  You know m to two powers, e1 = 3; e2 = 5,and you want the power 1.  How to combine these exponents and get 1?  sounds like xgcd in general: Hopefully 1 = a*e1 + b*e2  In this case it is easy: 
1 = 2*3 - 5;
m¹ = m^{2*3 - 5} = (m³) ²/ m ⁵ = (293) ²/ 421  in field Z/493Z

>>> from mod import Mod
>>> Mod(293, 493)**2/421
Mod(47, 493)
>>> pow(47, 3, 493) # check answer 47
293
>>> pow(47, 5, 493) # double check answer 47
421

Answer 47 

8.7.20: 
The book does not give Alice's public key, so I modified the problem statement to include it:  A = 26.  You still need her private key to decrypt.  Since p = 43 is small, we can solve the discrete log problem by brute force:
>>> from PohligHellman import discreteLogModP
>>> g = 3
>>> p = 43
>>> B = 30
>>> c = 7
>>> A = 26
>>> discreteLogModP(g, A, p)
17
>>> a = 17
>>> c*pow(B, p-1-a, p) % p # calc message
23
>>> m=23  # this is the answer
>>> discreteLogModP(g, B, p) # check - need B's private key
11
>>> b = 11
>>> m*pow(A, b, p) % p # confirm get c
7

Requested answer for the message was 23

15.3.2:
a(x) = 3 + 2x; get
x f(x)
1  5
2  7
3  9
4  0

Extra Review problem solutions

1&2.  Most elements of an elliptic group are paired, allowing (x, +y) and (x, -y), but the identity (point at infinity) is always an element, and it is not paired.  It is also possible for y to be 0, and (x, 0) is not paired, since +0 = -0.  A few tries for me with findE yielded a group with 6 elements.  Continuing I get 7 elements for problem 2.  I chose 7 as the modulus, since the number of group elements is on the order of the prime modulus:

Enter a prime modulus: 7
Enter polynomial modulus or just enter:
a b:1 3
6 elements:
0
(4, 6) (4, 1)
(5, 0)
(6, 6) (6, 1)
a b:3 5
7 elements:
0
(1, 3) (1, 4)
(4, 5) (4, 2)
(6, 6) (6, 1)

3. 21-1 = 20= 2*2*5   test powers d = 5 and 2*d
>>> a =13
>>> d = 5
>>> n = 21
>>> pow(a, d, n) # result neither 1 nor -1
13
>>> pow(a, 2*d, n) #result not -1
1

4.

1. A to B: 1037; B to A: 0; A to B: 1000;  1000**2 % n = 1037 OK
   
2. A to B: 381; B to A: 1; A to B: 434;  434**2 = 381*776 mod n OK
   
3. A to B: 691; B to A: 1; A to B: 733;  733**2 = 691*776 mod n OK
   
4. A to B: 711; B to A: 0; A to B: 304;  304**2 % n != 711  REJECT
   
5. A to B: 879; B to A: 1; A to B: 55;  55**2 != 879*776 mod n REJECT
   

5.  No.  The only connection to C is through A, and I have no reason to trust A.  The problem says who A trusts, not who trusts A.