Given that, $\sin x+\cos x=a$

On squaring both sides, we get

$$\begin{array}{rlrl} & (\sin x+\cos x)^2 =(a)^2 \\ \Rightarrow & \sin ^2 x+\cos ^2 x+2 \sin x \cos x =a^2 \\ \Rightarrow & \sin x \cdot \cos x =\frac{1}{2}\left(a^2-1\right) \end{array}$$

$$\begin{aligned} \text{(i) }\sin ^6 x+\cos ^6 x & =\left(\sin ^2 x\right)^3+\left(\cos ^2 x\right)^3 \\ & =\left(\sin ^2 x+\cos ^2 x\right)\left(\sin ^4 x-\sin ^2 x \cos ^2 x+\cos ^4 x\right) \\ & =\sin ^4 x+\cos ^4 x-\frac{1}{4}\left(a^2-1\right)^2 \\ & =\left(\sin ^2 x+\cos ^2 x\right)^2-2 \sin ^2 x \cos ^2 x-\frac{1}{4}\left(a^2-1\right)^2 \\ & =1-2 \cdot \frac{1}{4}\left(a^2-1\right)^2-\frac{1}{4}\left(a^2-1\right)^2=\frac{1}{4}\left[4-3\left(a^2-1\right)^2\right] \end{aligned}$$

$$\begin{aligned} \text{(ii) } |\sin x-\cos x| & =\sqrt{(\sin x-\cos x)^2} \\ & =\sqrt{\sin ^2 x+\cos ^2 x-2 \sin x \cos x} \\ & =\sqrt{1-2 \frac{1}{2}\left(a^2-1\right)}=\sqrt{1-a^2+1}=\sqrt{2-a^2} \end{aligned} $$

In right angled $\triangle A B C, \angle C=90 \Upsilon$

$$\begin{aligned} \therefore \quad \tan (A+B) & =\frac{\tan A+\tan B}{1-\tan A \tan B} \\ \Rightarrow \quad \frac{1}{0} & =\frac{\tan A+\tan B}{1-\tan A \tan B} \\ \Rightarrow \quad \tan A \tan B & =1 \quad \text{.... (i)}\\ \tan A+\tan B & =\frac{\sin A}{\cos A}+\frac{\sin B}{\cos B} \\ & =\frac{\sin A}{\cos A}+\frac{\sin (90 \Upsilon-A)}{\cos (90 \Upsilon-A)} \\ & =\frac{\sin A}{\cos A}+\frac{\cos A}{\sin A} \\ & =\frac{\sin ^2 A+\cos A}{\sin A \cdot \cos A} \\ & =\frac{1}{\sin A \cdot \cos A}=\frac{2}{2 \cdot \sin A \cdot \cos A} \\ & =\frac{2}{\sin 2 A} \quad [\because \sin2x=2\sin x \cos x] \end{aligned}$$

So, the required equation is $x^2-\left(\frac{2}{\sin A}\right) x+1$

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.