If $\gcd(a,n) = 1$ and $\gcd(b,n) = 1$, then $\gcd(ab,n) = 1$. If $\gcd(a,n) = 1$ and $\gcd(b,n) = 1$, then $\gcd(ab,n) = 1$.
Also... $a,b$ and $n$ are natural numbers. 
I feel I should begin with EEA to multiply out the gcd's, but I don't know where to go from there...
 A: Because $n$ is coprime to $a$ and $b$, we know there are linear combinations that equal one. Let the coefficients be $x,y,z,w$ such that $ax + ny = 1$ and $bz + nw = 1$. Multiply these together:
$$abxz + axnw + bzny + n^2yw = 1 \cdot 1$$
$$ab(xz) + n(axw + bzy + nyw) = 1$$
So there's a linear combination of $ab$ and $n$ that is equal to $1$. Thus they are coprime.
A: Hint: Suppose that $p$ is a prime divisor of $ab$ and of $n$; since $p$ is prime, it's necessarily true that $p \mid a$ or $p \mid b$. Can you take it from here?
A: In general gcd(a,n)=1 means there is no common divisor of a and n.. That is the composition multiples of a and n are different from each Other.. And gcd(b,n)=1 also gives the same result.. So now on multiplying a and b, there composition multiples will remain different from that of n. So gcd(ab,n)=1.. This is the simplest way to think about. 
A: There is a useful theorem for these kind of problems that says:
For $a,b \in \Bbb Z,a\neq 0, b\neq 0$ then $gcd(a,b)=1 \Leftrightarrow \exists m,n \in \Bbb Z : am+bn=1$
wich is not difficult to prove.
It is a useful exercise to prove or disprove this theorem when $gcd(a,b)=d$ and $am+bn=d$ where $d\ge 1$
