= Solution
Take the <inner product> with $u_i$ in $a=\sum_ja'_jw_j$. The <reciprocal basis> condition gives $a'_i=\langle u_i,a\rangle$. Expanding instead in the $u$ <basis> proves the <reciprocal-basis coordinate transformation>
$$
\boxed{a'_i=\sum_jG_{ij}a_j,\qquad G_{ij}=\langle u_i,u_j\rangle.}
$$
Thus $G$ is the <Gram matrix> of the original <basis>. It is a <positive-definite matrix>, so this <change of basis> is invertible.
In the specified two-dimensional Cartesian frame,
$$
u_1=(1,0),\qquad u_2=\left(\frac12,\frac{\sqrt3}{2}\right).
$$
Solving the four <dot product> conditions gives
$$
\boxed{w_1=\left(1,-\frac1{\sqrt3}\right),\qquad w_2=\left(0,\frac2{\sqrt3}\right).}
$$
They are perpendicular to $u_2$ and $u_1$, respectively, with lengths adjusted so the other <dot product> is one. Equating the Cartesian components of $a=a_1u_1+a_2u_2=a'_1w_1+a'_2w_2$ gives
$$
a_x=a_1+\frac12a_2=a'_1,\qquad
a_y=\frac{\sqrt3}{2}a_2=\frac{-a'_1+2a'_2}{\sqrt3}.
$$
Consequently
$$
\boxed{\begin{pmatrix}a'_1\\a'_2\end{pmatrix}=\begin{pmatrix}1&1/2\\1/2&1\end{pmatrix}\begin{pmatrix}a_1\\a_2\end{pmatrix},}
$$
which agrees entry by entry with the <Gram matrix> computed directly from the two unit <basis vectors>.
\Image[/past-exam-of-the-mathematical-methods-i-course-of-the-university-of-oxford/2018-reciprocal-bases.png]
{title=One <vector> reconstructed in two <reciprocal bases>: coordinates $(2,1)$ become $(5/2,2)$ through the <Gram matrix>}
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