What exactly interests you? Let me discuss $n=20.$

The number of primes below $2^{21}$ is evidently $155611$ of which $73586$ have exactly 20 bits. Given that, the answer to your question 2 with $n=20$ is definitely yes for $m$ below $73586$ and no for $m$ above $155611$. Doing question 1 for $m=70000$ would be tedious. If you wanted $n=20$ and $m=481$ then one could find a $k$ , a $20$ bit multiple of $2*3*5*7*11$ and take it along with $k+j$ for the $480$ numbers $j$ less than $k$ and relatively prime to it.

It might be more interesting to ask about finding $m$ numbers between $s$ and $t$ when $t-s$ is "small" with respect to $s.$ The case $s=2^n$ and $t=2^{n+1}$ seems too broad.

What exactly interests you? It might be more interesting to ask about finding $m$ pairwise relatively prime numbers between $s$ and $t$ when $t-s$ is "small" with respect to $s.$ The case $s=2^n$ and $t=2^{n+1}$ seems too broad. The best you can do is to take all the primes and prime powers in that range and augment that with some products $p\cdot p'$ where $p$ ranges over the primes less than $\sqrt{2^{n}}$ not used so far and, for each one, an appropriate prime cofactor $p'$.

Let me first discuss $n=20.$

The number of primes with exactly $20$ bits is $N=\pi(2^{21})-\pi(2^{20}) =73586$. Given that, the answer to your question 2 with $n=20$ is definitely yes for $m$ below $N+1$ Then you can use just primes. One can stuff in $234$ more using prime powers and semi-primes (detals below). Doing question 1 for $m=70000$ would be tedious. If you wanted $n=20$ and $m=481$ then one could find   $k$ , a $20$ bit multiple of $2\cdot 3\cdot 5\cdot 7\cdot 11=2310$ and take it along with $k+j$ for the $480$ numbers $j$ less than $2310$ and relatively prime to it.

What about finding the absolute largest size set of pairwise relatively prime $20$-bit integers? I claim that it is not hard to find that it is $N+57+5+6+166$ the extra terms are for

• $57$ prime squares, $p^2$ for $1331 \leq p \leq 1447$.
• $5$ prime cubes, $p^3$ for $p=103,107,109,113,127$
• $6$ more: $37^4 ,17^5 ,11^6, 5^9,3^{13}$ and $2^{20}.$
• $166$ primes under $1024$ not yet used (we will give each one a prime cofactor)

If we pair each of the $166$ small ones with the largest possible large prime we get

$7 \cdot 299569, 13\cdot 161309, 19\cdot 110359, 23\cdot 91163, 29\cdot 72313, 31\cdot 67631, 41\cdot 5113$

and continue on to $991\cdot 2113, 997 \cdot 2099, 1009 \cdot \mathbf{2069},1013 \cdot 2069,1019 \cdot \mathbf{2053},1021 \cdot 2053 $

The only overlaps are the two shown in bold which can be replaced with somewhat smaller primes such as $1997, 1999, 2003, 2011, 2017, 2027, 2029, 2039.$

The only "hard" step was perhaps that of computing $N=\pi(2^{21})-\pi(2^{20})=155611-82025=73586.$ Without that we wouldn't know exactly for what $m$ we pass from possible to not possible. But we would know how to get about where. For example easy estimates give $81519 \lt \pi(2^{20}) \lt 82158$ and $154701 \lt \pi(2^{21}) \lt 155852.$

Let $p_k$ be the $k$th prime and, for convenience, $p_0=1.$

For any non-negative $i$, the $m$ positive numbers $p_i,p_{i+1},p_{i+2},\cdots , p_{i+m-1}$ are pairwise relatively prime. Try $p_i$ the smallest prime with $n$ bits (or $1$ for $n=1$). If that fails then you should be able to do it for $n+1$ bits. That might be as good as it gets.

What exactly interests you? Let me discuss $n=20.$

The number of primes below $2^{21}$ is evidently $155611$ of which $73586$ have exactly 20 bits. Given that, the answer to your question 2 with $n=20$ is definitely yes for $m$ below $73586$ and no for $m$ above $155611$. Doing question 1 for $m=70000$ would be tedious. If you wanted $n=20$ and $m=481$ then one could find a $k$ , a $20$ bit multiple of $2*3*5*7*11$ and take it along with $k+j$ for the $480$ numbers $j$ less than $k$ and relatively prime to it.

It might be more interesting to ask about finding $m$ numbers between $s$ and $t$ when $t-s$ is "small" with respect to $s.$ The case $s=2^n$ and $t=2^{n+1}$ seems too broad.