Show there is exactly one ring homomorphism $g: R[x] \to S[x]$ s.t. $g(r)=f(r)$ give a ring homomorphism $f:R \to S$ and $p \in S[x]$
Show there is exactly one ring homomorphism $g: R[x] \to S[x]$ s.t. $g(r)=f(r), \forall r \in R$ and $g(x)=p$
So whatever $f$ does I'm not sure but seems for $r \in R$ a constant $g(r)$ takes $r$ to whatever $f(r)$ is in the first homomorphism. but I'm not sure from here. I think this one is probably not as  ugly as it looks.  
p is a polynomial in S[x] and g(x)=p.
so a constant goes to a constant and a non-constant polynomial goes to a specific polynomial p.
so $g(r_0+r_1x^1+r_2x^2+.....) = (f(r)+f(r_1)p+f(r_2)p^2+.......)$ I think.
EDIT:
its a ring homomorphism so 
$g((r_0+r_1x^1+r_2x^2+.....)+(r'_0+r'_1x^1+r'_2x^2+.....)) =(f(r_0+r'_0)+f(r_1+r'_1)p+f(r_2+r'_2)p^2+.......) = (f(r)+f(r_1)p+f(r_2)p^2+.......)+(f(r')+f(r'_1)p+f(r'_2)p^2+.......)= g((r_0+r_1x^1+r_2x^2+.....))+g((r'_0+r'_1x^1+r'_2x^2+.....))$
similar for multiplication but I'll wait to type that out until I get a better idea of what to do.
 A: The main property of the polynomial ring can be stated as follows.

Let $R$ and $B$ be (commutative) rings. If $f\colon R\to B$ is a ring homomorphism and $b\in B$, there is one and only one ring homomorphism $g_b\colon R[x]\to B$ such that
  
  
*
  
*$g_b(r)=f(r)$, for all $r\in R$
  
*$g_b(x)=b$
  

Uniqueness is easy: if $g_b$ is assumed to exist, then if $q(x)=r_0+r_1x+\dots+r_nx^n\in R[x]$, then
$$
g_b(q)=g_b(r_0)+g_b(r_1)g_b(x)+\dots+g_b(r_n)g_b(x)^n=
f(r_0)+f(r_1)b+\dots+f(r_n)b^n
$$
is completely determined by $f$ and $b$. It remains to do the (tedious but trivial) verification that defining
$$
g_b(r_0+r_1x+\dots+r_nx^n)=f(r_0)+f(r_1)b+\dots+f(r_n)b^n
$$
is indeed a ring homomorphism.
In your case, since $S$ is a subring of $S[x]$, we can think to $f$ as a ring homomorphism $f\colon R\to S[x]$. Now take $B=S[x]$ and $b=p(x)$ in the above setting.
A: Given $f : R \to S, \  p \in S[X]$, we can induce a ring hom $g: R[X] \to S[X]$ s.t. $g(r) = f(r),  \forall r \in R \subset R[X]$ and $g(X) = p(X)$.  Proof:
Define the hom in the obvious way by extending it linearly:
$$
g(a_0 + a_1 X + \dots a_n X^n) = g(a_0) + g(a_1) p(X) + g(a_2) p(X)^2 + \dots + g(a_n) p(X)^n
$$
Clearly this is a well-defined map as two equal polynomials have the same coefficients (formal polynomial equality, not function equality).
Now you must prove that $g$ is a homomorphism (tedious), and that it's unique (obvious or easy).
