Given integers $a,b,c,d\in[2^n,2^m]$ with $m>n>1$, how many primes $p$ are there in $[n^\alpha,n^\beta]$ for some $1<\alpha<\beta$ such that
$$0<a\bmod p<n^{\alpha/k}$$
$$0<b\bmod p<n^{\alpha/k}$$
$$0<c\bmod p<n^{\alpha/l}$$
$$0<d\bmod p<n^{\alpha/l}$$
holds where $k,l>2$ is fixed?

Assume $n,m,\alpha,\beta,k,l$ are fixed.


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Heuristically we can solve for average case of choice of $a,b,c,d$ following way:
$$\frac{|\{p:a\bmod p<n^{\alpha/k}\}|}{|\{p:p\in[n^\alpha,n^\beta]\}|}\approx\frac{|\{p:b\bmod p<n^{\alpha/k}\}|}{|\{p:p\in[n^\alpha,n^\beta]\}|}\approx\frac{n^{\alpha/k}}{(n^{\beta}-n^\alpha)/\log(n^{\beta}-n^\alpha)}$$
$$\frac{|\{p:c\bmod p<n^{\alpha/l}\}|}{|\{p:p\in[n^\alpha,n^\beta]\}|}\approx\frac{|\{p:d\bmod p<n^{\alpha/l}\}|}{|\{p:p\in[n^\alpha,n^\beta]\}|}\approx\frac{n^{\alpha/l}}{(n^{\beta}-n^\alpha)/\log(n^{\beta}-n^\alpha)}$$

So
$$\mathsf{Prob}(a\bmod p<n^{\alpha/k},b\bmod p<n^{\alpha/k},c\bmod p<n^{\alpha/l},d\bmod p<n^{\alpha/l})\underbrace{\approx}_{\mathsf{assuming}\mbox{ }\mathsf{independence}\mbox{ }\mathsf{of}\mbox{ }a,b,c,d}\frac{n^{2\alpha(1/k+1/l)}}{((n^{\beta}-n^\alpha)/\log(n^{\beta}-n^\alpha))^4}$$

So I think we should have approximately
$$\frac{n^{2\alpha(1/k+1/l)}}{((n^{\beta}-n^\alpha)/\log(n^{\beta}-n^\alpha))^3}\ll1$$ such prime numbers if $2\alpha(1/k+1/l)<3\beta$ which is if $\beta/\alpha>(1/k+1/l)/1.5$. Since $\beta/\alpha>1>(1/k+1/l)/1.5$ holds at every $k>1$ we have just $o(1)$ prime numbers.

Am I correct?

This is average case. 

Could there be as many as $\Omega(n^\gamma)$ special quadruples of $a,b,c,d$ such that each quadruple has $\Omega(n^\delta)$ of such primes for some $\gamma,\delta\in(0,1]$?