Solution to task 2:
$n=45$
The approach
Consider splitting the entire grid into cages, in this case the biggest $n$ we could hope to find is $81$ with a single cage which is obviously not going to work as it won't be solvable.
We need to get as close to this as possible while enforcing uniqueness, which means we need to first force some cell to be some value and then have that force another and so on cascading through the entire sudoku.
How to achieve that?
Use two cages and make it so one cage has rows and columns containing $(1),(1,2),\dots,(1,2,3,4,5,6,7,8,9)$ overlapping on the highest numbers and the other has the rest (overlapping on the smallest numbers). The two cages will now be the cages with the smallest and largest possible sums for their sizes too.
Two separate ones and two separate nines will fall out first, then two pairs of "a one and a two" and two pairs of "a nine and an eight", and so on until the whole sudoku is filled in the only way possible.
The cage sums will be
For the $[1,9]$ ($45$ cells) $\sum_{i=1}^9(\sum_{j=1}^ij)=165$
For the $[1,8]$ ($36$ cells) $405-165=240=\sum_{i=1}^8(\sum_{j=10-i}^9j)$
That is
$165=(1)+(1+2)+(1+2+3)+(1+2+3+4)+(1+2+3+5)+(1+2+3+4+5+6)+(1+2+3+4+5+6+7)+(1+2+3+4+5+6+7+8)+(1+2+3+4+5+6+7+8+9)$
And
$240=(9)+(9+8)+(9+8+7)+(9+8+7+6)+(9+8+7+6+5)+(9+8+7+6+5+4)+(9+8+7+6+5+4+3)+(9+8+7+6+5+4+3+2)$
Can we do it?
Yes, here is one:

which has the unique solution:

The first cells to fall out are:
$A9$ (the only column in $165$ with one cell);
$I1$ (the only row in $165$ with one cell);
$A8$ (the only row in $240$ with one cell); and
$F1$ (the only column in $240$ with one cell)
Now the rows and columns with two cells for each are uniquely defined - they must have values $(1,2)$ and $(9,8)$ respectively and one of each pair cannot be the $1$ or $9$ due to those already placed in either the same row or column - these are (in the same order as before):
$(D8,D9)$;
$(F2, I2)$;
$(A7,D7)$; and
$(C1,C2)$
This same process then cascades through to completion.
Now, as codewarrior0 notes, the smaller cage is unnecessary, as it is implicit, so we only need the $165$ cage of $45$ cells.