NP completeness of linear $0-1$ assignment problem
If I understood it well, (1) is also NP-complete, a possible reduction is from SUBSET SUM:

Given a set of $m$ positive integers $A = \{a_1, ..., a_m\}$, and a positive integer $B$, is there a subset of $A' \subseteq A$ a such that $\sum_{a_i \in A'} a_i$

You simply pick $n = (m+1)$ and build your set in this way:


add $ a_1, a_2, , ..., a_m$
add $m$ zeros
add the only negative number $-B$
"fill" the other remaining $n^2 - 2m - 1$ elements with $C = \sum_{a_i} + B +1$


($\Rightarrow$) If the original subset sum has a solution $A'$ then it is sufficient to use this partition $\Pi_1$

a1  C  C  ... C  0  C
 0 a2  C  ... C  C  C
        ....
 C  C     ... 0 am  C
 C  C     ... C  C -B


and for each "row" pick $a_i$ if $a_i \in A'$, $0$ otherwise (and pick the final $-B$)

($\Leftarrow$) If there is a partition $\Pi_1$ with a vanishing 0/1 assignment then note that:

(i)  none of the $C$s can be picked (they are too large);
(ii) at least one row must contain all elements different from $0$ (because there are $m+1$ rows and only $m$ zeros); so at least one nonzero element must be picked;
(iii) in every case the $-B$ element must be picked to vanish the sum

So the elements $a_i$ picked, no matter how the elements are partitioned between the rows, sum exactly to $B$; i.e. they are a valid solution for the original SUBSET SUM problem.

OUCH! I just read Michael's answer about problem (2) and the reduction is almost the same ... so you can also accept his answer!

EDIT: the reduction remains valid for the (1b) case:

The ($\Leftarrow$) direction it uses only the constraints on $\Pi_1$ which are the same of case 1a (exactly 1 "pick" from each row).

The ($\Rightarrow$) direction is also valid; to see it "visually", just build the main diagonal of $\Pi_1$ with the elements of $a_i \in A'$, the element $-B$, while the remaining elements of the diagonal are $0$s.

For example if $m = 5$ and the solution of the subset sum is $A' = \{ a_1, a_2, a_4 \}$, $\Pi_1$ is:

 a1  0  C   C  C   C
  C a2  0   C  C   C
  C  C  0  a3  C   C
  C  C  C  a4  0   C
  C  C  C   C  0  a5
  C  C  C   C  c  -B


In this way you can build $\Pi_2$ setting the element $j$ of row $i$-th equal to element $j$ of column $i$ in $\Pi_1$. The assignment clearly vanishes both $\Pi_1$ and $\Pi_2$ and the constraints are met.