= Solution
Let $\operatorname{ad}_A(B)=[A,B]$. Differentiating both exponentials and using that $A$ commutes with $e^{\lambda A}$ gives
$$
F'(\lambda)=Ae^{\lambda A}Be^{-\lambda A}-e^{\lambda A}Be^{-\lambda A}A
=e^{\lambda A}[A,B]e^{-\lambda A}.
$$
Applying the same calculation repeatedly gives $F^{(n)}(\lambda)=e^{\lambda A}\operatorname{ad}_A^n(B)e^{-\lambda A}$. Its <Taylor series> at zero, evaluated at one, proves the <commutator expansion for exponential conjugation>
$$
\boxed{e^ABe^{-A}=\sum_{n=0}^\infty\frac1{n!}\operatorname{ad}_A^n(B).}
$$
For bounded operators the series converges in <operator norm> by the bound $(2\|A\|)^n\|B\|$. The quantum computations below hold on a common invariant domain, such as Schwartz functions, with the <commutator> series either terminating or closing on a finite-dimensional span.
Use the <canonical commutation relations> $[\hat x,\hat p]=i\hbar$ and the <Heisenberg picture> $\hat B(t)=e^{itH/\hbar}\hat B e^{-itH/\hbar}$. For the first <Hamiltonian>,
$$
[H,\hat x]=-i\hbar\hat p,\qquad [H,\hat p]=-i\hbar\alpha,\qquad [H,[H,\hat x]]=-\hbar^2\alpha.
$$
All further relevant commutators vanish, so
$$
\boxed{\hat p(t)=\hat p+\alpha t,\qquad\hat x(t)=\hat x+t\hat p+\tfrac12\alpha t^2.}
$$
The explicit energy check is
$$
\frac12(\hat p+\alpha t)^2-\alpha(\hat x+t\hat p+\tfrac12\alpha t^2)
=\frac12\hat p^2-\alpha\hat x.
$$
The scalar shifts commute with all operators, so there is no ordering ambiguity.
For the two-dimensional system put $L=\hat x_1\hat p_2-\hat x_2\hat p_1$ and $V=-\beta L$. Direct commutation gives $[L,\hat p_1]=i\hbar\hat p_2$, $[L,\hat p_2]=-i\hbar\hat p_1$. Thus the <Heisenberg equations of motion> are $\dot{\hat p}_1=\beta\hat p_2$, $\dot{\hat p}_2=-\beta\hat p_1$, whose solution, equivalently the resummed <commutator> series, is
$$
\boxed{\hat p_1(t)=\hat p_1\cos\beta t+\hat p_2\sin\beta t,\qquad
\hat p_2(t)=-\hat p_1\sin\beta t+\hat p_2\cos\beta t.}
$$
Expanding both squares cancels their cross terms and gives $\hat p_1(t)^2+\hat p_2(t)^2=\hat p_1^2+\hat p_2^2$.
In the <interaction picture> generated by $H_0$, $[H_0,\hat x_i]=-i\hbar\hat p_i$ and $[H_0,\hat p_i]=0$. The terminating expansion therefore gives $\overline{x}_i(t)=\hat x_i+t\hat p_i$. Also $[H_0,L]=0$, either by the momentum-square calculation or by direct commutation. Conjugating $\overline{x}_i$ by $e^{itV/\hbar}$ rotates position and momentum by the same <matrix>, since $[L,\hat x_1]=i\hbar\hat x_2$ and $[L,\hat x_2]=-i\hbar\hat x_1$. Hence
$$
\boxed{\begin{aligned}
\hat x_1(t)&=(\hat x_1+t\hat p_1)\cos\beta t+(\hat x_2+t\hat p_2)\sin\beta t,\\
\hat x_2(t)&=-(\hat x_1+t\hat p_1)\sin\beta t+(\hat x_2+t\hat p_2)\cos\beta t.
\end{aligned}}
$$
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