Final solution
Step by step
First, look at the second cell in the bottom row, at the intersection of $11$ and $33$.
This is the largest number among three summing to $11$, so it must be one of $5,6,7,8$.
It is the smallest number among four summing to $33$, so it can't be $7$ or $8$.
If it is $5$, then above it must be $4$ and then $2$. In that case, consider the third cell in the middle row. It's bigger than $4$, so it must be at least $6$, but it's also the smallest number among three summing to $22$. Contradiction.
So this number is $6$, and above it is either $4$ then $1$ or $3$ then $2$, while the rest of the bottom row is either $12,8,7$ or $11,9,7$ or $10,9,8$.
Now consider the middle row and the third column.
The second cell in the middle row is either $3$ or $4$. The third cell in the middle row is bigger than that, but also the smallest number among three summing to $22$, and it's not $6$. So it must be either $4$ or $5$. If it's $4$, then the second column is $2,3,6$ (in order) and the middle row is $?,3,4,1$ summing to $22$, contradiction.
So the third cell in the middle row is $5$, and the other two must be either $8,9$ or $10,7$ in some order.
In the middle row, we have $5$ and two smaller numbers ($4,3$ or $4,2$ or $3,1$) and something else summing to $22$. So the first cell must be $10$ or $11$, and the $3,1$ option is impossible.
Assume the second cell in the middle row is $3$. Then the second column is $2,3,6$ (in order) and the middle row is $10,3,5,4$ (in order) and the third column is either $8,5,9$ or $9,5,8$. So the bottom row is either $7,6,8,12$ or $7,6,9,11$. But then the first column is $5,10,7$, contradiction.
So the second column is $1,4,6$ (in order), and we have:
If the last cell in the middle row is
$3$, then the middle row is $10,4,5,3$ (in order) so the other two cells in the first column must sum to $12$. But no possibilities are left for that, since $5,4,3,10,1$ are all used up. Contradiction.
So the last cell in the middle row is $2$, which means the middle row is $11,4,5,2$ in order, and again there's only one possibility for the first column:
Then the last four cells are $7,9,10,12$, with $7,12$ in one row and $9,10$ in the other. So the final solution must be: