Note that by Bezout's Theorem, there exist integers $s$ and $t$ such that $\gcd(a,b)=as+bt$. Similarly, there exist integers $u$ and $v$ such that $\gcd(a,c)=au+cv$.
Expand the product $(as+bt)(au+cv)$. We get an expression of the shape $ax+bcy$ for some integers $x$ and $y$.
Now if $e$ divides both $a$ and $bc$, then $e$ divides $ax+bcy$.
(1) The hypotheses $p|N$ and $q|N$ give us two integers $m,k$ so that $N=mp$ and $N=kq$. This implies $mp=kq$ so that $q|mp$. Now, since $gcd(p,q)=1$ and $q|mp$ we know that $q|m$ (think about why this is true if it's not clear). Then $m=sq$ for some integer $s$. Putting this all together, we have $N=mp=sqp=s(pq)$ so $pq|N$.
(2) Here we have an if and only if statement so you'll have to prove two statements/directions(or both at once):
(i) if $3|x$ then $3$ divides the sum of the digits of $x$
(ii) if $3$ divides the sum of the digits of $x$ then $3$ divides $x$
The strategy is to write the number in base $10$, for example (not a proof):
$1356=1\cdot 10^3+3\cdot 10^2+5\cdot 10^1 + 6\cdot 10^0$
Now $10$ has remainder $1$ under division by $3$ so the remainder of $1356$ under division by $3$ is $1 \cdot 1^3 + 3\cdot 1^2 +5\cdot 1^1 +6\cdot 1^0=1+3+5+6$ which is exactly the sum of the digits. Then the remainder under division by $3$ of $1356$ and $1+3+5+6$ are the same, and $3|1356$ if and only if $3|1+3+5+6$.
Try to do this in general for some number $n$ with logical steps following the idea of the example above to write a formal proof.
Best Answer
If $\gcd(a,b)=1$ then there exist integers $s$ and $t$ such that $a s +b t =1$. Hence $c a s +c b t =c$. Now since $a \vert c$ and $b \vert c$ there exist integers $m$ and $n$ such that $c = m a$ and $c= n b$. Hence $n b a s + m a b t =c$. Since $a b $ divides the entire left hand side, it must also divide the right hand side.