= Solution
Use $\omega_0^2=k/m$ and the power formula from <10/c/solution>. Its normalized form is
$$
\frac{\overline P(\omega)}{\overline P_{\max}}
=\frac{\gamma^2\omega^2}{(\omega_0^2-\omega^2)^2+\gamma^2\omega^2}.
$$
It is zero at $\omega=0$, rises to one at $\omega_0$, and decays to zero as $\omega\to\infty$. The two positive half-power <angular frequencies> satisfy
$$
(\omega_0^2-\omega^2)^2=\gamma^2\omega^2.
$$
On the low-frequency side, $\omega_0^2-\omega_1^2=\gamma\omega_1$; on the high-frequency side, $\omega_2^2-\omega_0^2=\gamma\omega_2$. Solving these <quadratic equations> gives
$$
\boxed{\omega_1=\frac{\sqrt{4\omega_0^2+\gamma^2}-\gamma}{2},\qquad
\omega_2=\frac{\sqrt{4\omega_0^2+\gamma^2}+\gamma}{2}.}
$$
Consequently the <frequency bandwidth of viscous oscillator power absorption> obeys the stronger exact result
$$
\boxed{\omega_2-\omega_1=\gamma.}
$$
For a sharp resonance, $\gamma\ll\omega_0$, expansion gives $\omega_{1,2}=\omega_0\mp\gamma/2+O(\gamma^2/\omega_0)$, which recovers the requested approximation. The exact width above assumes the constant viscous damping coefficient in this model. Although the curve is approximately symmetric near a sharp peak, the exact half-power points satisfy $\omega_1\omega_2=\omega_0^2$ rather than exact symmetry about $\omega_0$.
\Image[/past-exam-of-the-mathematical-methods-i-course-of-the-university-of-oxford/2019-oscillator-power-resonance.png]
{title=Normalized absorbed power for gamma divided by omega zero equal to 0.2. The peak and the two exact half-power frequencies are marked; their separation is gamma.}
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