= Solution
The real <matrix> $g=\begin{pmatrix}a&b\\c&d\end{pmatrix}$ with <determinant> one preserves $u\wedge v$. Hence $W_g(u)=W(gu)$ obeys the same <Weyl relations>. It is strongly continuous and irreducible because $g$ is invertible and its arguments range over the whole plane. Part (b) gives a unitary $V$ intertwining $W$ and $W_g$, so \b[$VW(u)V^*=W(gu)$]. If $V_1,V_2$ both work, $V_2^*V_1$ commutes with all $W$ and is scalar by part (a); its scalar has modulus one. Thus the implementer is unique up to phase.
For the displayed quarter-turn, choose the scaled inverse <Fourier transform>
$$
\boxed{(Vf)(s)=\pi^{-1/2}\int_{\mathbb R}e^{2ist}f(t)\,dt.}
$$
The formula starts on <Schwartz functions> and extends unitarily to $L^2$ by Plancherel and a change of scale. It conjugates $T_xf(t)=f(t+x)$ to multiplication by $e^{-2ixs}$, and multiplication by $e^{2iyt}$ to $T_y$. Keeping the Weyl phase therefore gives $VW(x,y)V^*=W(y,-x)$, as required. With the opposite Fourier-kernel sign one obtains the opposite quarter-turn, so that convention must be checked.
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