= Solution
With the convention in the question,
$$
T_r=\exp\left(\frac i\hbar r\cdot p\right)
$$
acts as
$$
(T_r\psi)(x)=\psi(x+r).
$$
Although $V(x+r)=V(x)$ for $r\in\Lambda$, the vector potential changes by
$$
A(x+r)=A(x)+\frac12B\times r.
$$
Consequently
$$
T_rHT_r^{-1}
=\frac1{2m}[p-eA(x+r)]^2+V(x)
\ne H
$$
for a nonzero magnetic field in general.
Now let $K=p+eA$. Since
$$
r\cdot K=-i\hbar r\cdot\nabla
+\frac e2r\cdot(B\times x)
$$
and the derivative in the $r$ direction annihilates
$r\cdot(B\times x)$, the two terms in its exponential commute. The <magnetic translation operator> therefore acts as
$$
\boxed{
(\mathcal T_r\psi)(x)
=\exp\left(\frac{ie}{2\hbar}
r\cdot(B\times x)\right)\psi(x+r)}.
$$
Part (a) gives $[K_i,p_j-eA_j]=0$, so $\mathcal T_r$ commutes with the kinetic term. Its phase commutes with $V$, while its translation sends $V(x)$ to $V(x+r)=V(x)$. Hence
$$
\boxed{[\mathcal T_r,H]=0\qquad(r\in\Lambda)}.
$$
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