Given that,

$$\begin{aligned} & \sec x \cos 5 x+1=0 \\ & \frac{\cos 5 x}{\cos x}+1=0 \Rightarrow \cos 5 x+\cos x=0 \end{aligned}$$

$\Rightarrow \quad 2 \cos \left(\frac{5 x+x}{2}\right) \cdot \cos \left(\frac{5 x-x}{2}\right)=0 \quad\left[\because \cos x+\cos y=2 \cos \frac{x+y}{2} \cdot \cos \frac{x-y}{2}\right]$

$$ \begin{array}{lrl} \Rightarrow & 2 \cos 3 x \cdot \cos 2 x & =0 \\ \Rightarrow & \cos 3 x & =0 \text { or } \cos 2 x=0 \\ \Rightarrow & \cos 3 x & =\cos \frac{\pi}{2} \text { or } \cos 2 x=\cos \frac{\pi}{2} \\ \therefore & 3 x & =\frac{\pi}{2} \Rightarrow 2 x=\frac{\pi}{2} \\ \text { and } & x & =\frac{\pi}{6} \Rightarrow x=\frac{\pi}{4} \end{array} $$ Hence, the solutions are $\frac{\pi}{2}, \frac{\pi}{4}$ and $\frac{\pi}{6}$.

Given that, $\sin(\theta+\alpha)=a\quad\text{... (i)}$

and $\sin(\theta+\beta)=b\quad \text{... (ii)}$

$$\begin{array}{ll} \therefore & \cos (\theta+\alpha)=\sqrt{1-a^2} \text { and } \cos (\theta+\beta)=\sqrt{1-b^2} \\ \therefore & \cos (\alpha-\beta)=\cos \{\theta+\alpha-(\theta+\beta)\} \end{array}$$

$$\begin{aligned} & =\cos (\theta+\beta) \cos (\theta+\alpha)+\sin (\theta+\alpha) \sin (\theta+\beta) \\ & =\sqrt{1-a^2} \sqrt{1-b^2}+a \cdot b=a b+\sqrt{\left(1-a^2\right)\left(1-b^2\right)} \\ & =a b+\sqrt{1-a^2-b^2+a^2 b^2} \end{aligned}$$

$$\begin{aligned} &\begin{aligned} \text{and}\quad & \cos (\alpha-\beta)=a b+\sqrt{1-a^2-b^2+a^2 b^2} \\ & =\cos 2(\alpha-\beta)-4 a b \cos (\alpha-\beta) \\ & =2 \cos ^2(\alpha-\beta)-1-4 a b \cos (\alpha-\beta) \\ & =2 \cos (\alpha-\beta)(\cos \alpha-\beta-2 a b)-1 \\ & =2\left(a b+\sqrt{1-a^2-b^2+a^2 b^2}\right)\left(a b+\sqrt{1-a^2-b^2+a^2 b^2}-2 a b\right)-1 \\ & =2\left[\left(\sqrt{1-a^2-b^2+a^2 b^2+a b}\right)\left(\sqrt{1-a^2-b^2+a^2 b^2}-a b\right)\right]-1 \\ & =2\left[1-a^2-b^2+a^2 b^2-a^2 b^2\right]-1 \\ & =2-2 a^2-2 b^2-1 \\ & =1-2 a^2-2 b^2\quad \text{Hence proved.} \end{aligned}\\ \end{aligned}$$

Given that, $$\cos (\theta+\phi)=m \cos (\theta-\phi)$$

$$\Rightarrow \quad \frac{\cos (\theta+\phi)}{\cos (\theta-\phi)}=\frac{m}{1}$$

$$\begin{aligned} &\text { Using componendo and dividendo rule, }\\ &\begin{aligned} \frac{\cos (\theta-\phi)-\cos (\theta+\phi)}{\cos (\theta-\phi)+\cos (\theta+\phi)} & =\frac{1-m}{1+m} \\ \Rightarrow \frac{-2 \sin \left(\frac{\theta-\phi+\theta+\phi}{2}\right) \cdot \sin \left(\frac{\theta-\phi-\theta-\phi}{2}\right)}{2 \cos \left(\frac{\theta-\phi+\theta+\phi}{2}\right) \cdot \cos \left(\frac{\theta-\phi-\theta-\phi}{2}\right)} & =\frac{1-m}{1+m} \\ \Rightarrow \quad \frac{\sin \theta \cdot \sin \phi}{\cos \theta \cdot \cos \phi} & =\frac{1-m}{1+m} \quad \left[\begin{array}{lr} \because \quad \sin (-\theta) & =-\sin \theta \\ \text { and } \cos (-\theta) & =\cos \theta \end{array}\right]\\ \Rightarrow \quad \tan \theta \cdot \tan \phi & =\frac{1-m}{1+m} \\ \Rightarrow \quad \tan \theta & =\left(\frac{1-m}{1+m}\right) \cot \phi \end{aligned} \end{aligned}$$

Given expression,

$$\begin{gathered} 3\left[\sin ^4\left(\frac{3 \pi}{2}-\alpha\right)+\sin ^4(3 \pi+\alpha)\right]-2\left[\sin ^6\left(\frac{\pi}{2}+\alpha\right)+\sin ^6(5 \pi-\alpha)\right] \\ \quad=3\left[\cos ^4 \alpha+\sin ^4(\pi+\alpha)\right]-2\left[\cos ^6 \alpha+\sin ^6(\pi-\alpha)\right] \\ \quad=3\left[\cos ^4 \alpha+\sin ^4 \alpha\right]-2\left[\cos ^6 \alpha+\sin ^6 \alpha\right]=3-2=1 \end{gathered}$$

Given that, $a \cos 2 \theta+b \sin 2 \theta=c$

$$\Rightarrow a\left(\frac{1-\tan ^2 \theta}{1+\tan ^2 \theta}\right)+b\left(\frac{2 \tan \theta}{1+\tan ^2 \theta}\right)=c \quad\left[\because \sin 2 \theta=\frac{2 \tan \theta}{1+\tan ^2 \theta} \text { and } \cos 2 \theta=\frac{1-\tan ^2 \theta}{1+\tan ^2 \theta}\right]$$

$$\begin{array}{ll} \Rightarrow & a\left(1-\tan ^2 \theta\right)+2 b \tan \theta=c\left(1+\tan ^2 \theta\right) \\ \Rightarrow & a-a-\tan ^2 \theta+2 b \tan \theta=c+c \tan ^2 \theta \\ \Rightarrow & (a+c) \tan ^2 \theta-2 b \tan \theta+c-a=0 \end{array}$$

Since, this equation has $\tan \alpha$ and $\tan \beta$ as its roots.

$$\because \quad \tan \alpha+\tan \beta=\frac{-(-2 b)}{a+c}=\frac{2 b}{a+c}$$