Prove that the remainder when $f(x)$ is divided by $x-a$ is $f(a)$. Let $F$ be a field and let $R=F[x]$. 
$1.$ Let $a\in F$ and $f(x)\in R.$ Prove that the remainder when $f(x)$ is divided by $x-a$ is $f(a).$
$2.$ Let $a\in F$ and $f(x) \in R.$ Prove that $x-a$ divides $f(x)$ if and only if $f(a)=0.$
For $1$, I'm thinking of considering a general polynomial and considering division by $x-a$. The statement is equivalent to saying that there is a polynomial $q(x)\in F[x]$ such that $f(x)=(x-a)q(x)+f(a)$.
For $2$, suppose I were to prove the equivalent statement for $1$. Then I assume that $f(a)=0$. This can only occur if $x-a | f(x)$ by the definition of divisibility. Hence the reverse implication is true. Now, if $(x-a) | f(x),$ then by the definition of divisibility, $\exists h(x) \in F[x]$ such that $f(x) = h(x)(x-a)$. However, we know that $f(x)$ can be written as $f(x)=(x-a)q(x)+f(a)\Rightarrow f(x)-f(a) = (x-a)q(x)\Rightarrow (x-a) | (f(x)-f(a))$. However, $(x-a) | f(x),$ so we must have that (this is easy to prove) $(x-a) | f(a)$. However, $f(a)=0$ or $\deg f(a) < \deg x-a=1,$ so we must have that $f(a)=0$ by the definition of polynomial division (I can probably expand this somehow).

Is the proof for the second question correct, assuming the proof for the first one is correct?

 A: In the forward direction, I'm not sure how you're claiming $f(a) = 0$ implies that $(x - a)\mid f(x).$ Of course, this is true, but the definition of divisibility is that $a\mid b$ if and only if there exists $c$ such that $ac = b.$ Why does $f(a) = 0$ produce a $g(x)$ such that $f(x) =(x-a)g(x)$? You should be more clear about this.
In the reverse direction, you correctly determine that if $(x-a)\mid f(x),$ then there exists $h$ such that $f(x) = (x-a)h(x).$ You eventually correctly obtain that either $f(a) = 0$ or $\deg f(a) < 1,$ but this simply means that $f(a)$ is a constant, which you already knew. If you want to use the definition of polynomial division at the end to conclude $f(a) = 0,$ you do need to include more detail. You can proceed as follows:

If $(x - a)\mid f(a),$ then we can write $f(a) = (x-a)\cdot g(x)$ for some $g\in F[x].$ However, if $f(a)\neq 0$, we have
  $$0 = \deg f(a) = \deg((x-a)g(x)) = \deg(x-a) + \deg(g(x)) = 1 + \deg(g(x))\geq 1,$$
  which is a contradiction. Thus, $f(a) = 0.$

However, once you obtain $f(x) = (x-a)h(x),$ there's a much easier way to proceed - plug in $x = a$!
