Subtask 1

The condition $A\subseteq X\subseteq B$ is equivalent to $(a\mathrm{\&}x)=a$ and $(b\mathrm{\&}x)=x$, which can be checked on $\mathcal{O}(1)$ time (here, $\mathrm{\&}$ denotes the bitwise AND operation). For each type $2$ query, we can loop through all $2^N$ subsets and add its value to the total if it satisfies this condition.

Time Complexity: $\mathcal{O}(Q\cdot 2^N)$

Subtask 2

If we treat the $2^N$ identifiers as vertices of an $N$-dimensional hypercube, we can create an $N$-dimensional prefix sum array in $\mathcal{O}(N\cdot 2^N)$ by calculating a prefix sum in each dimension. More specifically, for each $i$ $(0\le i<N)$, for all integers $x$ with $i^{th}$ bit $0$, add arr$[x]$ to arr$[x+2^i]$. Now arr$[a]$ will be the sum of the values of all $X$ such that $X\subseteq A$.

Split the queries into $\sqrt{Q}$ blocks of size $\sqrt{Q}$. At the start of each block, initialize the prefix sum array as above. Keep a list of the updates that have occurred, and when querying a set $A$, return arr$[a]$ with adjustments for any set $X\subseteq A$ whose value was updated (there are at most $\sqrt{Q}$ of them if you reset the list of updates each block).

Time Complexity: $\mathcal{O}((N\cdot 2^N+Q)\sqrt{Q})$

Subtask 3

Let $x[i]$ denote the $i^{th}$ bit of $x$. For each query $(A,B)$, we need $(a\mathrm{\&}b)=a$, otherwise the answer to the query would be $0$. This means there are $3$ possibilities for $(a[i],b[i])$: either $(0,0)$, $(0,1)$, or $(1,1)$. Let $x_1$ and $x_2$ be the first and second halves of the binary representation of an integer $x$. We can 'join' the $a_1$ and $b_1$ to get a ternary number $\text{join}(a_1,b_1)$, where the $i^{th}$ digit of $\text{join}(a_1,b_1)$ is $0$ if $(a_1[i],b_1[i])=(0,0)$, $1$ if $(a_1[i],b_1[i])=(0,1)$, or $2$ if $(a_1[i],b_1[i])=(1,1)$. Similarly, we can convert back, going from any $N/2$-digit ternary integer $c$ to a corresponding pair of $N/2$-bit integers $a$ and $b$, with $(a\mathrm{\&}b)=a$ and $\text{join}(a,b)=c$.

Now, define the array arr$[c][d_2]$ as the sum of the values of all $N$-bit integers $d$, whose second half (in binary) is $d_2$, and whose first half $d_1$ satisfies $(a\mathrm{\&}{d}_1)=a$ and $(b\mathrm{\&}{d}_1)=d_1$, where $a$ and $b$ are the pair of integers such that $(a\mathrm{\&}b)=a$ and $\text{join}(a,b)=c$. When the value of $s$ changes by $v$, we can update this array in $\mathcal{O}(2^{N/2})$ by adding $v$ to arr$[\text{join}(a,b)][s_2]$, for all $a,b$ such that $(a\mathrm{\&}{s}_1)=a$ and $(b\mathrm{\&}{s}_1)=s_1$ (there are exactly $2^{N/2}$ such pairs $a,b$). To answer the query $(a,b)$, we add up arr$[\text{join}(a_1,b_1)][x_2]$ for all $x_2$ such that $(a_2\mathrm{\&}{x}_2)=a_2$ and $(b_2\mathrm{\&}{x}_2)=x_2$, which is also $\mathcal{O}(2^{N/2})$. Note that we can precalculate $\text{join}(a,b)$ in $\mathcal{O}(N\cdot 3^{N/2})$, and array arr can be efficiently initialized in $\mathcal{O}(6^{N/2})$ (alternatively, you can initialize with all $0$'s and update $2^N$ times).

Time Complexity: $\mathcal{O}(Q\cdot 2^{N/2}+6^{N/2})$