= Solution
Let $\bar\omega=(\omega_++\omega_-)/2$ and $\delta=(\omega_+-\omega_-)/2$. The sum and difference identities give
$$
x_1=d\cos\bar\omega t\cos\delta t,\qquad
x_2=d\sin\bar\omega t\sin\delta t.
$$
For weak coupling, $\omega_+-\omega_-=\omega_-\epsilon+O(\epsilon^2)$, so the slow sine and cosine envelopes produce <beats>. The first maximum of the second envelope is at $t=\pi/(\omega_+-\omega_-)$, when the first envelope vanishes.
A literal <first turning point of an initially stationary coupled oscillator> is different from that slow envelope maximum. The first nonzero time at which the second mass is instantaneously at rest satisfies
$$
\boxed{\bar\omega\cos(\bar\omega t)\sin(\delta t)
+\delta\sin(\bar\omega t)\cos(\delta t)=0.}
$$
Set $z=\bar\omega t$ and $\eta=\delta/\bar\omega\ll1$. Dividing the derivative by $d\delta$ and expanding at $z$ of order one gives $z\cos z+\sin z+O(\eta^2)=0$. The first positive root lies in $(\pi/2,\pi)$, where $\tan z=-z$, so
$$
\boxed{t_{\rm first}=\frac{2.028757838\ldots}{\bar\omega}
+O\left(\frac{\eta^2}{\bar\omega}\right).}
$$
There is no earlier root: for $0<z\le\pi/2$, both terms in the exact derivative are positive when $\eta$ is sufficiently small. At the root the second displacement reaches its first local maximum. If the question instead intends the first \i[envelope] maximum, its time is $\pi/(\omega_+-\omega_-)\simeq\pi/(\epsilon\omega_-)$. That is a slow amplitude-transfer time, and is not generally a time of zero instantaneous velocity. The distinction matters because both masses keep oscillating while the envelopes change.
\Image[/past-exam-of-the-mathematical-methods-ii-course-of-the-university-of-oxford/2017-resit-coupled-oscillator-turning-and-beats.png]
{title=The first turning point and the much slower envelope transfer of coupled oscillators}
{description=Original dimensionless solutions at weak nonzero coupling. An early-time panel marks the first zero of the second velocity, while long-time curves show the separate beat envelopes.}
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