 Solution

The real matrix $g=\left(\begin{array}{cc}a& b\\ c& d\end{array}\right)$ 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 $VW(u)V^{\ast }=W(gu)$. If $V_1,V_2$ both work, $V_2^{\ast }{V}_1$ 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 transformThe formula starts on Schwartz functions and extends unitarily to $L^2$ by Plancherel and a change of scale. It conjugates $T_{x}f(t)=f(t+x)$ to multiplication by $e^{-2ixs}$, and multiplication by $e^{2iyt}$ to $T_y$. Keeping the Weyl phase therefore gives $VW(x,y)V^{\ast }=W(y,-x)$, as required. With the opposite Fourier-kernel sign one obtains the opposite quarter-turn, so that convention must be checked.