We have
$$ a_1−a_2−b_1+b_2=x $$
$$ a_1+a_2=n_1 $$
$$ b_1+b_2=n_2 $$
$$ a_1+b_1=n_1 $$
$$ a_2+b_2=n_2 $$
Noting that
$$a_2=n_1-a_1=b_1$$
and letting $a_2=k$, we see that we have
$$(a_1,a_2,b_1,b_2,x)=(n_1-k,k,k,n_2-k,n_1+n_2-4k)\tag1$$
for some $k$.
For 1., 2. :
The system has a solution in non-negative integers if and only if we have
$$n_1-k\ge 0\quad\text{and}\quad k\ge 0\quad\text{and}\quad n_2-k\ge 0\quad\text{and}\quad n_1+n_2-4k\ge 0\quad\text{where}\quad k\in\mathbb Z,$$
i.e.
$$0\le k\le \min\left\{n_1,n_2,\left\lfloor\frac{n_1+n_2}{4}\right\rfloor\right\}\quad\text{where}\quad k\in\mathbb Z$$
For 3., 4. :
From 1., 2.,
$$x_{\text{min}}=n_1+n_2-4\min\left\{n_1,n_2,\left\lfloor\frac{n_1+n_2}{4}\right\rfloor\right\}$$
For 5. :
Since $k=n_1/2$, from $(1)$,
$$(a_1,a_2,b_1,b_2,x)=\left(\frac{n_1}{2},\frac{n_1}{2},\frac{n_1}{2},\frac{n_1}{2},0\right)$$
is a solution.
Best Answer
$x^2+2y^2=11\tag{1}$
$[x,y]=[3,1]$ is a known solution for equation $(1)$.
Substitute $x=3+t, y=1+kt$ to equation $(1)$, then we get
$$t = \frac{-2(3+2k)}{1+2k^2}$$ Then we get a parametric solution.
$$x = (-3+6k^2-4k)/(1+2k^2)$$ $$y = -(-1+2k^2+6k)/(1+2k^2)$$ Thus, we get infinitely many rational solutions.
Example: $k=1..10$. $[x,y]=[1/3, 7/3], [13/9, 19/9], [39/19, 35/19], [7/3, 5/3], [127/51, 79/51], [189/73, 107/73], [263/99, 139/99], [349/129, 175/129], [447/163, 215/163], [557/201, 259/201]$