Mathematical field, show that 0·a = 0, (-1)·a = -a, ... Based on the axioms for a mathematical field, the wiki article states that 0·a = 0 and (-1)·a = -a are consequences of the axioms, but doesn't show how they are derived. There was a similar question asked before, but I'm not sure about the accepted answer.
https://en.wikipedia.org/wiki/Field_(mathematics)#Elementary_consequences_of_the_definition
Also, it would seem logical that if a ≠ b, and if c ≠ 0, then c·a ≠ c·b, a uniqueness property that should hold true for a finite field (unordered) that I'm wondering if it can be derived from the axioms (perhaps something like induction?) .
 A: For $0 \times a$:
$$
0 \times a = (0+0)\times a = 0\times a + 0 \times a .
$$
Now whatever $0 \times a$ is, it has an additive inverse, so you can subtract it from each side of that equation to conclude that $0 \times a = 0$.
For $(-1) \times a$:
$$
0 \times a = (1 + (-1)) \times a = 1 \times a + (-1) \times a = a + (-1) \times a
$$
but $a$ has a unique additive inverse $-a$.
For your third question. 
If $c \ne 0$ then it has a multiplicative inverse $d$. Then
$$
ca = cb \implies ca - cb = c(a-b) = 0 
\implies dc(a-b) = 0 \implies a-b = 0 
\implies a = b.
$$
A: I presume that by "-a" you mean the additive inverse of a so that you want to prove that -1 times a is the additive inverse of a.  Again, that follows from the distributive law.  (1+ (-1))a= 0a= 0.  But, by the distributive law, (1+ (-1)))a= 1a+ (-1)a= a+ (-1)a= 0 also.  "a+ (-1)a= 0" is precisely what is meant by "additive inverse of a".
The statement "if a ≠ b, and if c ≠ 0, then c·a ≠ c·b" is most easily proved by proving the "contrapositive".  For any statement "if p then q", the contrapositive is the statement "if not q then not p" and it is easy to show in general that a statement is true if and only if its contrapositive is true.
The contrapositive of "if a ≠ b, and if c ≠ 0, then c·a ≠ c·b" is "if ca= cb then a= b".  To prove that add the additive inverse of (subtract) cb from both sides: ca- cb= 0.  By the distributive law, c(a- b)= 0.  Since c is not 0 we must have a- b= 0 so a= b.
A: $(-1)a=-a $ means $a+(-1)a=0$ as additive inverses are unique.
$a+(-1)a=1*a+(-1)a $ (existence of multiplicative identity. 
$=(1+(-1))a $ (distributive property)
$=0*a $.  Remains to show $0*a=0$ for all $a $
$0*a= (0+0)*a $ (Associativity and definition of additive property.
$=0*a+0*a $.  
$0a=0a+0a $
$0a+(-0a)=0a+0a+(-0a) $ (existence of additive inverse and acknowledgement that addition is a closed binary opperation)
$0=0a+0=0a $(additive inverse and implied associtivity.)
So $a+(-1)a=0$.  So $(-1)a=-a$  (and $a= -(-1 (a)) $.)
