= Solution
A <Euclidean lattice> in $\mathbb R^2$ is a discrete additive <subgroup>, equivalently the <integer> span of zero, one or two real-linearly independent <vectors>. Its rank is the number of these generators, or the dimension of its real span. Discreteness makes its intersection with any bounded set finite, so shortest nonzero vectors exist.
First $\Lambda\cap\mathbb Rw_1=\mathbb Zw_1$. Otherwise an element $tw_1$ with noninteger $t$ could be reduced by its nearest <integer> multiple of $w_1$ to a nonzero vector of length at most $|w_1|/2$, contradicting minimality. Therefore $w_2\notin\mathbb Zw_1$ is not parallel to $w_1$.
Write any $v\in\Lambda$ as $v=aw_1+bw_2$ with real coefficients. Subtract nearest <integer> multiples to obtain $r=sw_1+tw_2\in\Lambda$, with $|s|,|t|\le1/2$. If $t\ne0$, it is not on the first axis, yet
$$
|r|\le\tfrac12(|w_1|+|w_2|)\le|w_2|,
$$
with strict inequality because the independent vectors cannot achieve equality in the <triangle inequality>; if one coefficient vanishes the inequality is also strict. This contradicts minimality of $w_2$. If $t=0$ and $r\ne0$, then $|r|\le|w_1|/2$, contradicting minimality of $w_1$. Thus $r=0$ and both original coefficients are <integers>. Hence \b[$\boxed{\Lambda=\mathbb Zw_1+\mathbb Zw_2}$].
For $w_1=(1,0)$ and $w_2=(x,y)$, minimality implies $x^2+y^2\ge1$ and $|w_2|\le|w_2\pm w_1|$, hence $|x|\le1/2$. The exact region is
$$
\boxed{-\tfrac12\le x\le\tfrac12,\qquad |y|\ge\sqrt{1-x^2}.}
$$
These conditions are sufficient: in the generated lattice, vectors with second coefficient $\pm1$ have smallest length when the first coefficient is zero, allowing ties at $|x|=1/2$. A vector with second coefficient of magnitude at least two has length at least $2|y|>\sqrt{x^2+y^2}$, since $|y|^2\ge3/4$ and $x^2\le1/4$. Vectors on the first axis have length at least one. Thus the shaded region includes exactly all allowable choices, including ties on its boundary.
\Image[/past-exam-of-the-mathematics-course-of-the-university-of-cambridge/2011/ii/paper-4-lattice-basis.png]
{title=Allowed second shortest lattice vectors in the vertical strip outside the unit circle}
For a second lattice basis, write $v_1=aw_1+bw_2$, $v_2=cw_1+dw_2$ with <integer> coefficients, as just proved. Conversely each $w_i$ is an <integer> combination of the $v_j$. The change-of-basis matrix $\begin{pmatrix}a&c\\b&d\end{pmatrix}$ and its inverse therefore both have <integer> entries. Their <determinants> are <integers> with product one, giving \b[$\boxed{ad-bc=\pm1}$]. Thus lattice bases are related by a <unimodular matrix>.
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