Putting $a=2^d$ and $k=\frac{m}{d}, l=\frac{n}{d}$ we have - in (a) $gcd(2^m-1,2^{2n}-1)=gcd(a^k-1,a^{2l}-1)=a^{gcd(k,2l)}-1=a-1$ but $gcd(a-1,a^l+1)=1$ so indeed $gcd(2^m-1,2^n+1)=gcd(a^k-1,a^l+1)=1$ so we are done - in (b) $gcd(2^m-1,2^{2n}-1)=gcd(a^k-1,a^{2l}-1)=a^{gcd(k,2l)}-1=a^2-1$ (cause $l$ is odd and $k$ is even), so for sure $gcd(a^k-1,a^l+1)|a^2-1$, and it's easy to see that $a+1$ divides both $a^k-1$ and $a^l+1$ and also that $a-1$ does not divide $a^l+1$ - if not $a-1=2^d-1|2$ (but for $2^d-1|2$ we have $d=1$ and $a-1=1$ so that doesn't change our gcd) thus $gcd(2^m-1,2^n+1)=a+1=2^d+1$ .

Can you plz explain $gcd(a^k-1,a^{2l}-1)=a^{gcd(k,2l)}-1$???

Rushil wrote: Can you plz explain $gcd(a^k-1,a^{2l}-1)=a^{gcd(k,2l)}-1$??? Because $gcd(r,s)=gcd(r-s,s)$. $gcd(a^x-1,a^y-1)=gcd(a^x-a^y,a^y-1)$. Because $gcd(a,a^y-1)=1$, $gcd(a^x-a^y,a^y-1)=gcd(a^{x-y}-1,a^y-1)=gcd(a^{x-2y}-1,a^y-1)=...=gcd(a^{x mod y}-1,a^y-1)$. Repeating this the exponents will be the residues in Euclidean algorithm, so $gcd(a^x-1,a^y-1)=a^{gcd(x,y)}-1$. Of course only for a>1.