= Solution
In the scalar two-particle model, the <fermionic exchange symmetry> and <bosonic exchange symmetry> conditions are
$$
\boxed{\Psi(x_2,x_1)=-\Psi(x_1,x_2)\text{ for fermions},\qquad
\Psi(x_2,x_1)=\Psi(x_1,x_2)\text{ for bosons}.}
$$
For particles with spin, exchange acts on all their labels; the stated coordinate-only model fixes the spatial symmetry sector. Define $S\Psi(x_1,x_2)=\Psi(x_2,x_1)$. Equal masses make the kinetic operator invariant under exchange, and the assumed potential is invariant too, so $HS=SH$. Therefore $H\Psi=E\Psi$ implies $H(S\Psi)=E(S\Psi)$. The components
$$
\Psi_{\mathrm B}=\tfrac12(\Psi+S\Psi),\qquad
\Psi_{\mathrm F}=\tfrac12(\Psi-S\Psi)
$$
are respectively symmetric and antisymmetric eigenfunctions whenever nonzero. Their sum is the original nonzero <quantum state>, so at least one is nonzero. An eigenspace may contain both sectors if the energy is degenerate.
The <normal modes of two harmonically coupled identical particles> provide the separation. The <coordinate transformation> to $y_+,y_-$ is <orthogonal>, with absolute Jacobian one. Exchanging the particles leaves $y_+$ fixed and changes $y_-$ to $-y_-$. Hence \b[bosonic <quantum states> are even and fermionic <quantum states> odd in the relative coordinate]. The <chain rule> gives
$$
\partial_{x_1}=\frac{\partial_{y_+}+\partial_{y_-}}{\sqrt2},\qquad
\partial_{x_2}=\frac{\partial_{y_+}-\partial_{y_-}}{\sqrt2},\qquad
\partial_{x_1}^2+\partial_{x_2}^2=\partial_{y_+}^2+\partial_{y_-}^2.
$$
Also $x_1^2+x_2^2=y_+^2+y_-^2$ and $(x_1-x_2)^2=2y_-^2$. Combining the two trapping potentials with the interaction therefore gives the separated <Hamiltonian operator>
$$
H=\left[-\frac{\hbar^2}{2m}\partial_{y_+}^2+\frac12m\omega^2y_+^2\right]
+\left[-\frac{\hbar^2}{2m}\partial_{y_-}^2+\frac12m\omega^2\cos^2\alpha\,y_-^2\right].
$$
These are independent oscillators of frequencies $\omega$ and $\omega\cos\alpha$. Both frequencies are positive in the stated range. Their product <eigenstates> have energies
$$
\boxed{E_{n_+,n_-}=\hbar\omega\left[n_++\tfrac12+(n_-+\tfrac12)\cos\alpha\right],\qquad n_+,n_-=0,1,2,\ldots.}
$$
The <parity> of an oscillator level is $(-1)^n$. Consequently <bosons> allow even $n_-$ and <fermions> odd $n_-$, while $n_+$ is unrestricted.
The bosonic <ground state> uses $n_+=n_-=0$, and the fermionic <ground state> uses $n_+=0,n_-=1$. The oscillator length in the relative mode is
$$
\boxed{\tau^2=\frac{\hbar}{m\omega\cos\alpha}=\frac{\hbar\sec\alpha}{m\omega}.}
$$
Using the normalized <Gaussian> for each mode gives
$$
\boxed{\Phi_{\mathrm B}(y_+,y_-)=\psi_\sigma(y_+)\psi_\tau(y_-),\qquad
\Phi_{\mathrm F}(y_+,y_-)=\psi_\sigma(y_+)\phi_\tau(y_-).}
$$
Indeed $\phi_\tau(y)=\sqrt2\,y\psi_\tau(y)/\tau$ by differentiating the <Gaussian>, so it is the normalized odd first <excited state>. The ground energies are $E_{\mathrm B}=\hbar\omega(1+\cos\alpha)/2$ and $E_{\mathrm F}=\hbar\omega(1+3\cos\alpha)/2$.
Choose normalized factors for a normalized product <quantum state>; this can always be done by reciprocal rescaling of its factors. Since the coordinate Jacobian has absolute value one,
$$
\mathbb E[f(y_-)]=\iint f(y_-)\,|\psi(y_+)|^2|\phi(y_-)|^2\,dy_+dy_-
=\boxed{\int f(y)|\phi(y)|^2\,dy.}
$$
If the supplied factors are not individually normalized, the general expression has an additional factor $\|\psi\|^2$; choosing normalized factors is the convention in the displayed formula.
For the <ground states>, put $\rho(y)=|\psi_\tau(y)|^2=e^{-y^2/\tau^2}/(\sqrt\pi\tau)$. Then $|\phi_\tau(y)|^2=2y^2\rho(y)/\tau^2$ and $\rho'=-2y\rho/\tau^2$. Thus, for differentiable $f$ with suitable <Gaussian> integrability and vanishing boundary terms,
$$
\mathbb E_{\mathrm F}[f]=\frac2{\tau^2}\int y^2f(y)\rho(y)\,dy
=-\int yf(y)\rho'(y)\,dy
=\boxed{\int\bigl(f(y)+yf'(y)\bigr)\rho(y)\,dy.}
$$
This is <integration by parts>, and every polynomial satisfies the required endpoint conditions. The <relative-coordinate moments in even and odd oscillator ground states> follow by substituting $f(y)=y^n$:
$$
\boxed{\mathbb E_{\mathrm F}[y_-^n]=(n+1)\mathbb E_{\mathrm B}[y_-^n].}
$$
For odd $n$, both expectations vanish by <parity>, so this is an equality of zeros rather than an increase in signed mean separation. The even moments show the physical difference. In particular $\mathbb E_{\mathrm B}[y_-^2]=\tau^2/2$ and $\mathbb E_{\mathrm F}[y_-^2]=3\tau^2/2$, so the fermionic root-mean-square particle separation is $\sqrt3$ times the bosonic value. The fermionic <probability density> vanishes at coincidence $y_-=0$, expressing <Pauli exclusion principle> in this spatial symmetry sector. This is an exchange constraint on the <wave function>, not an additional classical force. As $\alpha$ approaches $\pi/2$, the relative oscillator softens and its width diverges; the excluded endpoint would no longer have this normalizable oscillator <ground state>.
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