= Solution
{c}
A nondegenerate <hyperbola> centred at the origin has a <quadratic equation>
$$
Ax^2+2Bxy+Cy^2+F=0.
$$
There are no linear terms because the origin is its centre. Its <asymptotes> are the zero set of the homogeneous <quadratic form> $Ax^2+2Bxy+Cy^2$. Requiring the directions $y=x$ and $y=-x$ gives $A+2B+C=A-2B+C=0$, hence $B=0$ and $C=-A$. Since the <conic section> is nondegenerate, $A$ and $F$ are nonzero. Thus all such <hyperbolas with prescribed asymptotes> are exactly
$$
\boxed{H_c:\ x^2-y^2=c,\qquad c\in\mathbb R\setminus\{0\}.}
$$
For $c>0$ these <rectangular hyperbolas> open along the $x$-axis, and for $c<0$ along the $y$-axis. Conversely, each displayed <hyperbola> has the required centre and <asymptotes>. Two distinct members have different values of $x^2-y^2$, so \b[they cannot intersect].
For the second pair of <asymptotes>, the same argument gives $H_2:4x^2-y^2=d$, $d\ne0$. The <intersections of concentric hyperbolas> therefore satisfy
$$
x^2=\frac{d-c}{3},\qquad y^2=\frac{d-4c}{3}.
$$
If both right sides are positive, there are four independent sign choices. If precisely one is zero and the other is positive, there are two points. If either is negative there are none; both cannot vanish because $c\ne0$. Each possibility occurs: take $(c,d)=(1,5)$ for four, $(1,4)$ for two, and $(1,1)$ for none. Hence
$$
\boxed{|H_1\cap H_2|\in\{0,2,4\}.}
$$
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