Answer
$$\boxed{ \theta=\frac{1}{2}\tan^{-1}\left[\frac{g}{a}\right]}$$
Work Step by Step
We need to know that the initial (and final) horizontal velocity component of the cannon's projectile relative to the ground is given by
$$v_{x}=v_0\cos\theta+v_{\rm train}\tag 1$$
since the train and the cannon are moving in the same direction.
And the vertical velocity component of the cannon's projectile relative to the ground is given by
$$v_{iy}=v_0\sin\theta \tag 2$$
The final vertical velocity component of the cannon's projectile relative to the ground is given by
$$v_{fy}=v_0\sin\theta -gt\tag 3$$
since it moves under free-fall acceleration.
Noting that in equation (3), $v_{fy}=0$ at the highest point of the projectile's trip.
This means that the time needed for its half trip is
$$t_1=\dfrac{v_0\sin\theta }{g}$$
Hence, the time needed for the whole trip is given by
$$t_{tot}=2t_1=\dfrac{2v_0\sin\theta }{g}\tag 4$$
Now we know that the distance between the falling point of the projectile and the train is the difference between the distance traveled by the projectile and the distance traveled by the train.
See the figure below for more clarification.
Thus,
$$ x_{\rm diff}=x_{\rm projectile}-x_{\rm train}\tag 5$$
We know that the distance traveled by the projectile is given by
$$x_{\rm projectile}=v_x t_{tot}$$
Plugging from (1) and (4);
$$x_{\rm projectile}=(v_0\cos\theta+v_{\rm train})\dfrac{2v_0\sin\theta }{g} $$
$$x_{\rm projectile}= \dfrac{2v_0^2\sin\theta \cos\theta}{g}+\dfrac{2v_0v_{\rm train}\sin\theta }{g} \tag 6 $$
And the distance traveled by the train is given by
$$x_{train}=v_{\rm train}t_{tot}+\frac{1}{2}at_{tot}^2$$
Plugging from (4);
$$x_{train}=\dfrac{2v_0v_{\rm train}\sin\theta}{g}+ \dfrac{a(2v_0\sin\theta)^2}{2g^2}$$
$$x_{train}=\dfrac{2v_0v_{\rm train}\sin\theta}{g}+ \dfrac{ 2v_0^2a\sin^2\theta }{ g^2}\tag 7$$
Plug (6) and (7) into (5); and note the signs.
$$x_{\rm diff}=\dfrac{2v_0^2\sin\theta \cos\theta}{g}+\dfrac{2v_0v_{\rm train}\sin\theta }{g}-\dfrac{2v_0v_{\rm train}\sin\theta}{g}- \dfrac{ 2v_0^2a\sin^2\theta }{ g^2} $$
The second and the third terms on the right side are having the same magnitude with opposite signs, so they are canceled.
$$x_{\rm diff}=\dfrac{2v_0^2\sin\theta \cos\theta}{g} - \dfrac{ 2v_0^2a\sin^2\theta^2}{ g^2} $$
$$x_{\rm diff}=\dfrac{2v_0^2}{g}\left[\sin\theta \cos\theta-\dfrac{ a\sin^2\theta }{ g } \right] $$
Since we need to maximize the distance $\Delta x$, so we need $$\dfrac{dx_{\rm diff}}{d\theta}=0$$
$$\dfrac{d}{d\theta}x_{\rm diff}=\dfrac{2v_0^2}{g}\dfrac{d}{d\theta}\left[\sin\theta \cos\theta-\dfrac{ a\sin^2\theta }{ g } \right]=0 $$
$$\dfrac{d}{d\theta}\left[\sin\theta \cos\theta-\dfrac{ a\sin^2\theta }{ g } \right]$$
$$\cos\theta^2-\sin\theta^2-\dfrac{a}{g}(\sin\theta\cos\theta+\sin\theta\cos\theta)=0 $$
$$\overbrace{\cos\theta^2-\sin\theta^2}^{\cos2\theta}-\dfrac{a}{g}(\overbrace{2\sin\theta\cos\theta}^{\sin2\theta}) =0 $$
$$\cos2\theta-\frac{a}{g}\sin2\theta=0$$
Thus,
$$\cos2\theta=\frac{a}{g}\sin2\theta $$
$$\dfrac{\sin2\theta}{\cos2\theta}=\frac{g}{a}=\tan2\theta$$
Hence,
$$2\theta=\tan^{-1}\left[\frac{g}{a}\right]$$
$$\boxed{ \theta=\frac{1}{2}\tan^{-1}\left[\frac{g}{a}\right]}$$
$\bullet$ If the train was speeding up, so $a\gt 0$ and hence, $\theta\lt 45^\circ$ since $2\theta$ lies in the first quadrant.
$\bullet$ If the train was slowing down, so $a\lt 0$ and hence, $\theta\gt 45^\circ$ since $2\theta$ lies in the second quadrant.
$\bullet$ If the train was not accelerating at all, so $a= 0$ and hence, $\theta= 45^\circ$ since $2\theta=90^\circ$.
