Show that $\gcd(a,bc)=1$ if and only if $\gcd(a,b)=1$ and $\gcd(a,c)=1$

One direction should be clear: If $a$ and $bc$ have no common factors greater than $1$, then certainly neither do $a$ and $b$ nor $a$ and $c$. If this is not quite clear, argue thus: If $q$ divides $c$, then $q$ divides $bc$. So, any common factor of $a$ and $c$ is also a common factor of $a$ and $bc$. Same with $b$ in place of $c$.

For the other direction, Bezout's lemma gives a nice argument. Note that if $m,n$ are integers, and there are integers $x$ and $y$ such that $mx+ny=1$, then $\mathrm{gcd}(m,n)=1$. This is because any common divisor of $m,n$, is also a divisor of $mx+ny$. The other direction also holds, and is the content of Bezout's lemma: If $\mathrm{gcd}(m,n)=1$, then there are integers $x,y$ with $mx+ny=1$.

Accordingly, fix integers $x,y,z,w$ such that $ax+by=1$ and $az+cw=1$. Multiply these two equations, to obtain $$ a(axz+xcw+byz)+bc(yw)=1. $$ The numbers $k=axz+xcw+byz$ and $l=yw$ are integers, and we have $ak+(bc)l=1$, so (as explained above), it follows that $\mathrm{gcd}(a,bc)=1$.

By the way, this idea of multiplying linear combinations given by Bezout allows us to prove many similar results. For example, if $\mathrm{gcd}(a,b)=1$, then also $\mathrm{gcd}(a^2,b^3)=1$, etc. (Several problems along these lines have been asked on the site before.)

For the part that if $\gcd(a,b)=1$ and $\gcd(a,c)=1$, then $\gcd(a,bc)=1$:

Hint:

If $1=ax+by=au+cv$, then

$$1=(ax+by)(au+cv)=a(aux+cvx+byu)+bc(yv)$$