Given expression, $3(\sin x-\cos x)^4+6(\sin x+\cos x)^2+4\left(\sin ^6 x+\cos ^6 x\right)$

$$\begin{aligned} &= 3\left[\sin ^2 x+\cos ^2 x-2 \sin x \cos x\right]^2+6\left[\sin ^2 x+\cos ^2 x+2 \cdot \sin x \cdot \cos x\right] \\ & \quad+4\left[\left(\sin ^2 x\right)^3+\left(\cos ^2 x\right)^3\right] \\ &= 3(1-\sin 2 x)^2+6(1+\sin 2 x)+4\left[\left(\sin ^2+\cos ^2 x\right)\left(\sin ^4 x-\sin x \cos ^2 x+\cos ^4 x\right)\right. \\ &=3\left(1+\sin ^2 23 x-2 \sin 2 x\right)+6+6 \sin 2 x+4\left[\left(\sin ^2 x+\cos ^2 x\right)^2 3 \sin x \cos ^2 x\right] \\ &=3+3 \sin ^2 2 x-6 \sin 2 x+6+6 \sin 2 x \\ &= 4-3 \sin ^2 2 x=13 \end{aligned}$$

Given function, $f(x)=-3 \cos \sqrt{3+x+x^2}$

We know that,

$$\begin{aligned} -1 \leq \cos x \leq 1 \\ \Rightarrow \quad -3 \leq 3 \cos x \leq 3 \\ \Rightarrow \quad 3 \geq-3 \cos x \geq-3 \\ \Rightarrow \quad -3 \leq-3 \cos x \leq 3 \end{aligned}$$

So, the value of $f(x)$ lies in $[-3,3]$.

Given that, $y=\sqrt{3} \sin x+\cos x$

$$\begin{aligned} y & =2\left[\frac{\sqrt{3}}{2} \sin x+\frac{1}{2} \cos x\right] \\ & =2\left[\sin x \cdot \cos \frac{\pi}{6}+\cos x \sin \frac{\pi}{6}\right] \\ & =2 \sin (x+\pi / 6) \end{aligned}$$

Graph of $y=2\sin x$

Hence, the maximum distance is 2 units.