= Solution
For an odd integer $x=2a+1$, $x^2=1+4a(a+1)\equiv1\pmod8$. The same conclusion holds for every <2-adic unit> by reduction modulo eight. Thus an odd square must be congruent to one modulo eight.
Conversely let $u\in\mathbb Z_2^\times$ with $u\equiv1\pmod8$. For <digit lifting for square roots of 2-adic units>, start with $x_3=1$. If $x_n$ is odd and $x_n^2\equiv u\pmod{2^n}$, consider
$$
x_{n+1}=x_n+\varepsilon_n2^{n-1},\qquad\varepsilon_n\in\{0,1\}.
$$
For $n\ge3$,
$$
x_{n+1}^2\equiv x_n^2+\varepsilon_n2^nx_n\pmod{2^{n+1}},
$$
since the remaining term is divisible by $2^{2n-2}$ and $2n-2\ge n+1$. The odd multiplier $x_n$ lets the choice of $\varepsilon_n$ correct exactly the next binary digit. Thus $x_{n+1}^2\equiv u\pmod{2^{n+1}}$. The corrections have valuations tending to infinity, so completeness gives $x=\lim x_n\in\mathbb Z_2^\times$ with $x^2=u$.
The unit $3$ is congruent to one modulo two but is not a square, and $5$ is congruent to one modulo four but is not a square, by the necessary modulo-eight condition. Therefore
$$
\boxed{\text{The smallest }k\text{ is }3,\qquad u\text{ is a square}\ \Longleftrightarrow\ u\equiv1\pmod8.}
$$
Every nonzero <P-adic number> in $\mathbb Q_2$ is uniquely $2^m u$ with $m\in\mathbb Z$ and $u$ a <2-adic unit>. A square has even valuation, and the unit square criterion shows that the unit class is determined by its odd residue modulo eight. The four residues form $C_2\times C_2$, generated by $-1$ and $5$; the valuation parity supplies the third factor, generated by $2$. Hence the <square-class group of the 2-adic numbers> is
$$
\boxed{\mathbb Q_2^\times/(\mathbb Q_2^\times)^2\cong C_2^3,\qquad\text{basis }[-1],[2],[5].}
$$
Its eight classes are represented by $1,-1,5,-5,2,-2,10,-10$. They are distinct: the first four have even valuation and distinct unit residues modulo eight, while multiplication by $2$ gives odd valuation and the same four distinct unit residues.
In characteristic different from two, completing the square in a quadratic minimal polynomial makes every quadratic extension $K(\sqrt d)$ for a nonsquare $d\in K$. If $K(\sqrt d)=K(\sqrt e)$, write $\sqrt e=a+b\sqrt d$ and square. The coefficient $2ab$ must vanish. Since $e$ is nonsquare, $b\ne0$, so $a=0$ and $e/d=b^2$. Conversely a square ratio plainly gives the same field. Thus the seven <quadratic extensions of the 2-adic numbers> are exactly
$$
\boxed{\mathbb Q_2(\sqrt{-1}),\ \mathbb Q_2(\sqrt5),\ \mathbb Q_2(\sqrt{-5}),\ \mathbb Q_2(\sqrt2),\ \mathbb Q_2(\sqrt{-2}),\ \mathbb Q_2(\sqrt{10}),\ \mathbb Q_2(\sqrt{-10}).}
$$
The unique <unramified extension> in this list is $\mathbb Q_2(\sqrt5)$. To verify this, $\alpha=(1+\sqrt5)/2$ is integral and satisfies $X^2-X-1=0$. Its reduction $X^2+X+1$ is irreducible over $\mathbb F_2$, so its residue field contains $\mathbb F_4$. The standard degree formula for finite extensions of complete discretely valued fields with perfect residue field is $[L:K]=e(L/K)f(L/K)$. Here the field degree is two, so the <residue degree> is two and the <ramification index> is one.
Conversely an unramified quadratic extension has residue field $\mathbb F_4$. A root there of $X^2+X+1$ lifts by <Hensel lemma> to a root of $X^2-X-1$, whose derivative is a unit. For this lift $\beta$, $(2\beta-1)^2=5$, so the extension contains $\mathbb Q_2(\sqrt5)$ and must equal it. Every other field in the list is therefore ramified.
For the requested <ramification groups>, choose $L=\mathbb Q_2(\sqrt2)$ and $\pi=\sqrt2$. We use uniqueness of the extension of the valuation to a finite extension, together with the degree formula just stated. Since $v_2(\pi)=1/2$, the extension has ramification index two and residue degree one. Normalize $v_L(\pi)=1$, giving $v_L(2)=2$. Every element is $a+b\pi$ with $a,b\in\mathbb Q_2$, and the two summands have valuations of opposite parity:
$$
v_L(a+b\pi)=\min\{2v_2(a),\ 2v_2(b)+1\}.
$$
Thus $a+b\pi$ is integral exactly when $a,b\in\mathbb Z_2$, proving $\mathcal O_L=\mathbb Z_2\oplus\mathbb Z_2\pi$ directly.
The nonidentity automorphism $\sigma$ negates $\pi$. For any integral $a+b\pi$,
$$
\sigma(a+b\pi)-(a+b\pi)=-2b\pi,
$$
of valuation at least three, with equality when $b$ is a unit. By the definition $G_i=\{\sigma:v_L(\sigma x-x)\ge i+1\text{ for every }x\in\mathcal O_L\}$, the <ramification in the quadratic extension generated by the square root of two> is therefore
$$
\boxed{G_1=G_2=\operatorname{Gal}(L/\mathbb Q_2)\cong C_2,\qquad G_i=1\text{ for every }i\ge3.}
$$
The lower ramification break is two; the valuation used here is the integral normalization on $L$.
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