= Solution
For $a,b\in\mathbb Q_p^\times$, the <quadratic Hilbert symbol> $(a,b)_p$ is one if $b$ is a <field norm> from $\mathbb Q_p(\sqrt a)$, and minus one otherwise; a square $a$ gives the trivial extension and symbol one. It depends only on the two <square classes>. Equivalently, the conic $z^2=ax^2+by^2$ has a nonzero <rational point> precisely when the symbol is one. Interchanging $a$ and $b$ in the conic gives $(a,b)_p=(b,a)_p$. The <Hilbert norm residue symbol> is a <group homomorphism> in each argument; for nonsquare $a$, the <cyclic local norm index> makes its <group kernel> a <subgroup> of index two. The resulting pairing on <square classes> is nondegenerate. Also
$$
(a,-a)_p=1,\qquad(a,a)_p=(a,-1)_p,
$$
since $N(\sqrt a)=-a$. In the language of <local class field theory>, $(a,b)_p=\operatorname{Art}_p(b)(\sqrt a)/\sqrt a$; in the <Brauer group>, it records whether the quaternion <cyclic algebra> $(a,b)$ splits. These are compatible norm-obstruction descriptions.
An <unramified extension> of <local fields> of degree two has norm <group> consisting precisely of elements with even <valuation>. Indeed,
$$
v_K(Nz)=2v_L(z),
$$
and the norms on units are surjective. For the latter assertion, the norm on the finite <residue fields> is surjective. On successive <higher principal-unit groups>, the congruence
$$
N(1+\pi^r t)\equiv1+\pi^r\operatorname{Tr}_{\kappa_L/\kappa_K}(\bar t)\pmod{\pi^{r+1}}
$$
allows one to correct a proposed unit norm one digit at a time: the residue <field trace> is surjective, and <completeness> makes the corrections converge. An unramified quadratic field is unique up to <isomorphism>. Therefore, if $d$ is its defining nonsquare class,
$$
\boxed{(d,b)_p=(-1)^{v_p(b)}.}
$$
For odd $p$, choose the nonsquare unit $u_0$ from part (i). The <polynomial> $T^2-u_0$ has irreducible separable reduction, so it defines the unramified <quadratic extension>. Thus
$$
\Delta_1=\{1,u_0\},\qquad\Delta_2=\{1,u_0\}.
$$
The trivial class is included in $\Delta_2$, since a degree-one extension is unramified. The two units pair trivially, while $(u_0,p)_p=-1$. Any class outside $\Delta_1$ has odd <valuation> and consequently pairs nontrivially with $u_0$. This proves directly
$$
\boxed{\Delta_1^\perp=\Delta_2,\qquad\Delta_2^\perp=\Delta_1\quad(p\ne2).}
$$
For $p=2$, the unit <square classes> from part (i) give $\Delta_1=\{1,-1,5,-5\}$. The element $\omega=(1+\sqrt5)/2$ satisfies $T^2-T-1$; its reduction $T^2+T+1$ is irreducible and separable over $\mathbb F_2$. This shows that $\mathbb Q_2(\sqrt5)$ is unramified of degree two, giving
$$
\Delta_1=\{1,-1,5,-5\},\qquad\Delta_2=\{1,5\}.
$$
For a fully explicit pairing check, use the generators $(-1,5,2)$. We have $(-1,5)_2=1$ and $(-1,2)_2=1$, since $5=N(1+2i)$ and $2=N(1+i)$ from $\mathbb Q_2(i)$. But $(-1,-1)_2=-1$: a norm equal to $-1$ would be a sum of two squares. If either summand had negative <valuation>, the sum would have negative <valuation> (equal negative <valuations> give <valuation> $1-2r<0$); if both were integral, their squares could not sum to $-1\equiv3\pmod4$. Finally, the unramified norm criterion gives $(5,2)_2=-1$. The identity $(a,b)_2=(b,a)_2$ and $(a,a)_2=(a,-1)_2$ determine the remaining entries. Writing $(a,b)_2=(-1)^{B(a,b)}$, the exponent <matrix> is
$$
B=\begin{pmatrix}1&0&0\\0&0&1\\0&1&0\end{pmatrix}
\quad\text{in the basis }(-1,5,2).
$$
For $b=(-1)^s5^t2^r$, annihilating both generators $-1,5$ of $\Delta_1$ requires $s=r=0$, so $b\in\{1,5\}$. Annihilating $5$ alone requires $r=0$, so $b$ is a unit class. Hence
$$
\boxed{\Delta_1^\perp=\Delta_2,\qquad\Delta_2^\perp=\Delta_1\quad(p=2).}
$$
These are the <Hilbert-symbol annihilators of units and unramified square classes>. Annihilators here are taken with respect to this pairing, or equivalently in the <character group of a finite abelian group> after identifying <square classes> with their Hilbert-symbol characters.
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