If $a| x^2$ does that mean that $a$ will also always divide $x$? Also if $x^2$ has a remainder $b$ when divided by $a$ could you prove that $x$ also has a remainder b when divided by $a$ ?

Both are not true. Take x as 6 and a as 9. Then a divides $x^2$ but a does not divide x. In the second case take x as 4 and a as 5. 16 leaves a remainder 1 when divided by 5, but 4 leaves a remainder 4 when divided by 5. Think of a condition when what you say can be true, at least the first statement.

no, forexample let $a=x^2=16$ then $a$ divide $x^2$ but doesnot divide $4$

If $x$ divided by $a$ leaves remainder $b$, then the remainder when dividing $x^2$ by $a$ is $b^2$. Check modular arithmetic for discussion.

That is false in general: $4$ divides $36=6^2$, but $4\nmid 6$.

It is true if $a$ is prime, because of Euclid's lemma:

If a prime number divides a product, it divides one of its factors.

For your second question, I don't see what you mean: all integers can be divided and have a remainder $(0$ or not).