(a) $m$ divides $m^{2n}.$ Now we can observe that now if we divide $m^{2n}$ by $m+1$ then remainder will always be $1.$ Now to make remainder $0$ we have to subtract $1$ from $m^{2n}$ so in that way we can say that $m+1$ divides $m^{2n}-1.$
(b) It is popular theorem by Mathematician Fermat
https://en.wikipedia.org/wiki/Fermat%27s_little_theorem
(c) To make $mx+ny=1,$ $n$ and $m$ both must have to be relatively prime. If only one is prime and then other, when it is a multiple of that prime, sum cannot be 1.
like $5x+ny=1$ if $n$ is $10,15,20\ldots$ result cannot be $1.$ Basically the theorem is
For any positive integers $a$ and $b$, there exist integers $x$ and $y$ such that
$mx+ny= gcd(m, n).$
Now when $m$ and $n$ are relatively prime then their $\gcd$ will be $1$ so eventually equation would be like $mx+ny=1$
(d) We know that ${}^nC_m$ means choose $m$ out of $n$ and result of this will always be integer. If we see carefully what's being asked is ${}^nC_m,$ which is always an integer so yes $m!$ will have to divide $n\times (n-1)(n-2)\ldots (n-m+1)$.
(e) https://en.wikipedia.org/wiki/Mersenne_prime
So, option C must be false.