= Solution
The perturbations obey $\delta g_{00}=2a^2\psi$, $\delta g_{0i}=-a^2B_i$, and $\delta g_{ij}=2a^2(\phi\delta_{ij}-E_{ij})$. Apply part (i). The time-time component changes by $-2a^2(T'+\mathcal HT)$, and the time-space component changes by $a^2(L_i'-\partial_iT)$. Hence
$$
\boxed{\widetilde\psi=\psi-T'-\mathcal HT,\qquad
\widetilde B_i=B_i+\partial_iT-L_i'}.
$$
For the spatial components,
$$
\delta\widetilde g_{ij}-\delta g_{ij}
=2a^2\left[\mathcal HT\delta_{ij}+\partial_{(i}L_{j)}\right].
$$
Taking the trace and the symmetric trace-free part separately gives
$$
\boxed{\widetilde\phi=\phi+\mathcal HT+\tfrac13\partial_iL^i,\qquad
\widetilde E_{ij}=E_{ij}-\partial_{\langle i}L_{j\rangle}},
$$
where $\partial_{\langle i}L_{j\rangle}=\frac12(\partial_iL_j+\partial_jL_i)-\delta_{ij}\partial_kL^k/3$. This explicitly accounts for the trace term in $\phi$; it must not be absorbed into the trace-free shear.
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