Using explicit Bezout identities obscures the essence of the matter. Since $\,d = \gcd(a,b)\,$ divides every integral linear combination of $\,a,b,\,$ we deduce that $\,d\mid 21,10,\,$ so $\,d\mid\gcd(21,10) = 1.$
The final inference can be deduced by either Bezout: $\,1 = \gcd(21,10) = 21 j + 10k\,$ for $\,j,k\in\Bbb Z,\,$ thus $\,d\mid 21,10\,\Rightarrow\,d\mid 1,\,$ or, better, it can be deduce immediately from the universal property of the gcd: $\,d\mid m,n\iff d\mid \gcd(m,n)$.
In order to apply the divisibility rule, you have to distinguish between odd and even positioned digits. (It doesn't matter how you count.)
Example:
In 77 the first position is odd, the second even, so you would have to calculate $7-7=0$, which is divisible by 11.
Now it should be easy for you to understand what you try to prove in (b): If a,b,c are three digits abc is the number $100a+10b+c$. You know what it means to say that this number is divisible by 11. You have to prove that $$11\vert (a+c)-b \Leftrightarrow 11\vert 100a+10b+c$$ or with modular arithmetic
$$ (a+c)-b \equiv 0 \pmod{11}\Leftrightarrow 100a+10b+c\equiv 0 \pmod {11}\; .$$
I don't want to spoil the fun so I leave it there.
P.S. Sorry, I hadn't noticed the answer posted in the meantime.
Best Answer
It is true that $(4!)^n\mid(4n)!$, but it is not obvious (at least, not to me). You can use the fact that, for any natural number $m$, $4!\mid m(m+1)(m+2)(m+3)$; after all$$\frac{m(m+1)(m+2)(m+3)}{4!}=\binom{m+3}4.$$So:
and therefore $(4!)^n\mid(4n)!$.