For your first question, the easiest way to do it is to note that primes $p=8k+1$ correspond to split primes of $\mathbb Q(\sqrt{-1},\sqrt{2})$, and since a density $1$ set of primes is split, it's equivalent to ask for the fraction of primes $\mathfrak p$ of $\mathbb Q(\sqrt{-1},\sqrt{2})$ such that $1-\sqrt{2}$ is a quadratic residue mod $\mathfrak p$. This density is $1/2$ unless $1-\sqrt{2}$ is a square in $\mathbb Q(\sqrt{-1},\sqrt{2})$.

It is not a square, as if it were equal to $(a+ib)^2$ for $a,b\in \mathbb Q(\sqrt{2})$, we would have $2ab=0$ so $a=0$ or $b=0$ and it would equal either $a^2$ (always positive) or $-b^2$ (always negative), but $1+\sqrt{2}$ is sometimes positive and sometimes negative.

So the answer is always yes.

For your second question, the answer you seek is the distribution of the number of roots of the equation $ ((y^2-c)/b )^2 = a$. When this equation has four roots, then $x^2=a$ has two roots, both with $bx+c$ quadratic residues. When it has no roots, then $x^2=a$ has two roots, one with $bx+c$ quadratic residue and one not. The density of the remaining set, where $x^2=a$ has two roots, neither a quadratic residue, can be found by subtracting $1/2$ from the proceeding equation.

So you want to apply Chebotarev to the Galois group of this equation. So for instance if the Galois group is full $D_4$, as should happen for generic $a,b,c$, the probability of two quadratic residues is $1/8$, one quadratic residue is $1/4$, two roots of $x^2=a$ but no quadratic residues is $1/8$.

However the Galois group is not always very large. If $x^2=a$, then $(dx+e)^2 = 2de x+ (e^2+ d^2a)$, so if we happen to have $b=2de$, $c=e^2+d^2a$ then all the roots are quadratic residues.