A.
Definition of independence:  P and Q are independent just in case Pr(Q|P) = Pr (Q)
NOTE: if  Pr(Q|P) = Pr(Q), then Pr(P|Q) = Pr(P); that is, if Q is independent of P, then P is independent of Q.
Proof:
Pr(P&Q) = Pr(P) x Pr(Q|P).  Hence, Pr(Q|P) = [Pr(P&Q)/Pr(P)].  Suppose now that  Q is independent of P, that is,  Pr(Q|P) = Pr(Q).  Then,  [Pr(P&Q)/Pr(P)] = Pr(Q).  Multiplying both sides by Pr(P)/Pr(Q), one gets  [Pr(P&Q)/Pr(Q)] = Pr(P).  But [Pr(P&Q)/Pr(Q)] = Pr(P|Q).  Hence, Pr(P|Q) = Pr(P).

B.
Proof of Bayes' Theorem
We know that Pr(P&Q) = Pr(P) x Pr(Q|P), and Pr(Q&P) = Pr(Q) x Pr(P|Q).  But (P&Q) and (Q&P) are logically equivalent, an consequently their probability is the same.  So,
Pr(P) x Pr(Q|P) = Pr(Q) x Pr(P|Q).  Dividing both sides by Pr(P), one gets  Pr(Q|P) =  Pr(P|Q) x [Pr(Q)/Pr(P)], which is Bayes' Theorem.

C.
A more extended version of Bayes' Theorem
Since P is logically equivalent to [(P&Q) v (P&-Q)] (check it out!), Pr(P) = Pr{[(P&Q) v (P&-Q)]}.  But since (P&Q) and (P&-Q) are mutually exclusive, Pr{[(P&Q) v (P&-Q)]} = Pr(P&Q) + Pr (P&-Q).
So, Pr(P) = Pr(P&Q) + Pr (P&-Q).  But  Pr(P&Q) = Pr(Q) x Pr(P|Q), and Pr (P&-Q) = Pr(-Q) x Pr(P|-Q).  So,  Pr(P) =  Pr(Q) x Pr(P|Q) + Pr(-Q) x Pr(P|-Q).  Hence,  by substituting the last equality in Bayes' Theorem, one gets:

Pr(P|Q) x Pr(Q)
Pr(Q|P) =   -------------------------------------------
Pr(Q) x Pr(Q|P) + Pr(-Q) x Pr(P|-Q)

D.
Pr(A&B) is less or equal to Pr(A)
Proof:
The proof is a reductio.  Suppose that  Pr(A&B) > Pr(A).  Then, Pr(A&B) > Pr[(A&B) v (A&-B)].  But since (A&B) and (A&-B) are mutually exclusive,
Pr(A&B) > Pr(A&B) + Pr(A&-B).  Simplifying, one gets 0 >  Pr(A&-B), which is imposible because no probability is less than 0.

E.
Pr(AvB) is greater or equal to Pr(A)
Proof:
The proof is a reductio.  Suppose that Pr(A) > Pr(AvB).  Then, Pr(A) > Pr(A) + Pr(B) - Pr(A&B).  Hence, Pr(A&B) > Pr(B), which is impossible.

F.
The Logical Consequence Principle: If A entails B (i.e., if "A; hence B" is valid), then Pr(B) is not less than Pr(A)
Proof:
Suupose that A entails B.  Then -A v B is a tautology.  (WHY?)  Hence, Pr(-A v B) = 1.  So, Pr(-A) + Pr(B) - Pr(-A&B) = 1.  Consequently, 1-Pr(A) + PrB) - Pr(-A&B) = 1.
So, Pr(B) - Pr(A) = Pr(-A&B).  But Pr(-A&B) is not less than 0.  Hence, Pr(B) is not less than Pr(A).

G.
A bet on P is fair just in case W/L = [1-Pr(P)]/Pr(P).
Proof:
A bet on P is fair just in case the betting quotient equal Pr(P), that is Pr(P) = L/(L+W); that is, L Pr(P) + W Pr(P) = L; that is, W Pr(P) = L - L Pr(P); that is,  WPr(P) = L(1-Pr(P)); so,
W/L = [1-Pr(P)]/Pr(P) .