As Bill Dubuque likes to insist, use the universal properties. Recall that $d=\gcd(a,b)$ if and only if:
- $d|a$ and $d|b$; and
- If $c|a$ and $c|b$, then $c|d$.
Using this property, we can proceed:
$\gcd(a,b)|a,b$, so $c\gcd(a,b)|ac,ab$, so $c\gcd(a,b) |\gcd(ac,bc)$. Conversely, $c|\gcd(ac,bc)$, so $\frac{1}{c}\gcd(ac,bc)|\frac{1}{c}(ac),\frac{1}{c}(bc)$; hence $\frac{1}{c}\gcd(ac,bc)|a,b$, so $\frac{1}{c}\gcd(ac,bc)|\gcd(a,b)$. Hence $\gcd(ac,bc)|c\gcd(a,b)$. Thus, $c\gcd(a,b)=\gcd(ac,bc)$.
$\def\lcm{\mathrm{lcm}}$We can use the dual of the gcd, the lcm. $m=\lcm(a,b)$ if and only if:
- $a|m$ and $b|m$; and
- If $a|k$ and $b|k$, then $m|k$.
Since $\gcd(a,b)\lcm(a,b) = |ab|$, we have: $\begin{align*} a|c\text{ and }b|c &\iff \lcm(a,b)|c\\ &\iff \frac{|ab|}{\gcd(a,b)}|c\\ &\iff |ab||\gcd(a,b)c\\ &\iff ab|\gcd(a,b)c. \end{align*}$ In particular, if $\gcd(a,b)=1$, then $ab|c$.
Or, if you don't want to use the lcm (or the formula $\gcd(a,b)\lcm(a,b)=|ab|$), note that $ab|c$ if and only if $\gcd(ab,c)=ab$. Since $a|c$ we can write $c=ak$, so \gcd(ab,c) = \gcd(ab,ak) = a\gcd(b,k)$.
Since $b|c = ak$, and $\gcd(a,b)=1$, then $b|k$, so we can write $k=b\ell$. So \gcd(ab,c) = \gcd(ab,ak) = a\gcd(b,k) = a\gcd(b,b\ell) = ab\gcd(1,\ell) = ab.$ Therefore, ab|c.
Number 3 is incorrect as currently stated. The fact that you can find some x$, $y$, and $z$ such that $\gcd(a,b,c) = ax+by+cz$, follows from the fact that $\gcd(a,b,c)=\gcd(\gcd(a,b),c)$ (check the universal property), and that $\gcd(a,b)$ can be expressed as $ak+b\ell$ for some $k,\ell\in\mathbb{Z}$.