For the first problem,
clearly, we need to eliminate $\theta,\phi$
Now, $\cos\phi=\frac{\cos\alpha}{\cos\beta}$ and $\cos\theta=\frac{\cos\alpha}{\cos\gamma}$
$$\text{and }\sin^2\alpha=\left(2\sin\dfrac\phi2\cdot\sin\dfrac\theta2\right)^2=(2\sin^2\dfrac\phi2)(2\sin^2\dfrac\theta2)=(1-\cos\phi)(1-\cos\theta)$$
$$\implies 1-\cos^2\alpha=\left(1-\frac{\cos\alpha}{\cos\beta}\right)\left(1-\frac{\cos\alpha}{\cos\gamma}\right)$$
$$\implies \cos^2\alpha(1+\cos\beta\cos\gamma)=\cos\alpha(\cos\beta+\cos\gamma)$$
Assuming $\cos\alpha\ne0, \cos\alpha=\frac{\cos\beta+\cos\gamma}{1+\cos\beta\cos\gamma}$
Applying Componendo and dividendo, $$\frac{1-\cos\alpha}{1+\cos\alpha}=\frac{1+\cos\beta\cos\gamma-(\cos\beta+\cos\gamma)}{1+\cos\beta\cos\gamma+\cos\beta+\cos\gamma}=\frac{(1-\cos\beta)(1-\cos\gamma)}{(1+\cos\beta)(1+\cos\gamma)}$$
As $\cos2A=\frac{1-\tan^2A}{1+\tan^2A}, \tan^2A=\frac{1-\cos2A}{1+\cos2A}$
$$\implies \tan^2\frac\alpha2=\tan^2\frac\beta2\cdot\tan^2\frac\gamma2 $$
For the second question, $$\tan\alpha= \tan^3\frac\theta2$$
$$\implies \frac{\sin\alpha}{\sin^3\frac\theta2}=\frac{\cos \alpha}{\cos^3\frac\theta2}=\pm\sqrt{\frac{\sin^2\alpha+\cos^2\alpha}{\left(\sin^3\frac\theta2\right)^2+\left(\cos^3\frac\theta2\right)^2}}$$
$$\text{Now,} \left(\sin^3\frac\theta2\right)^2+\left(\cos^3\frac\theta2\right)^2=\left(\sin^2\frac\theta2\right)^3+\left(\cos^2\frac\theta2\right)^3$$
$$=\left(\frac{1-\cos\theta}2\right)^3+\left(\frac{1+\cos\theta}2\right)^3 \text{ as }\cos2A=1-2\sin^2=2\cos^2A-1$$
$$=\frac{2(1+3\cos^2\theta)}8=\frac{m^2}4 \text{ as }\cos^2\theta=\frac{m^2-1}3$$
$$\implies \frac{\sin\alpha}{\sin^3\frac\theta2}=\frac{\cos \alpha}{\cos^3\frac\theta2}=\pm \frac2m$$
$$\implies \sin^3\frac\theta2=\pm\frac{m\sin\alpha}2\implies \sin\frac\theta2=\left(\pm\frac{m\sin\alpha}2\right)^{\frac13}$$
Similarly, find $\cos \frac\theta2$ and use $\cos^2\frac\theta2+\sin^2\frac\theta2=1$ to eliminate $\theta$
If you solve the equations for $(x,y)$ and play with some trigonometric identities, there is another (more symmetric but equivalent) parametrization
$$x=\frac{1}{4} (10 \cos (t)-5 \cos (3 t)-\cos (5 t))$$
$$y=\frac{1}{4} (10 \sin (t)+5 \sin (3 t)-\sin (5 t))$$
Using the multiple angle formulae, this reduces to
$$x=\cos(t)\,(5-4\cos^4(t))$$
$$y=\sin(t)\,(5-4\sin^4(t))$$
This makes
$$(x+y)^2=(\cos (t)+\sin (t))^{10} \qquad \text{and} \qquad (x-y)^2=(\cos (t)-\sin (t))^{10}$$
I am sure that, from here, you can finish.
Best Answer
By sum to product and product to sum formulas we have
$$\begin{cases} \sin \theta + \sin \phi = 2\sin\left(\frac{\theta+\phi}2\right)\cos\left(\frac{\theta-\phi}2\right)=x\\ \cos \theta + \cos \phi = 2\cos\left(\frac{\theta+\phi}2\right)\cos\left(\frac{\theta-\phi}2\right)=y \\ \tan \frac {\theta}{2} \tan \frac {\phi}{2} =\frac{\cos\left(\frac{\theta-\phi}2\right)-\cos\left(\frac{\theta+\phi}2\right)}{\cos\left(\frac{\theta-\phi}2\right)+\cos\left(\frac{\theta+\phi}2\right)} =z \end{cases} \implies\begin{cases} 2ab=x\\ 2cb=y\\ \frac{b-c}{b+c}=z \end{cases}$$
and since $a^2+c^2=1$ we obtain
and then
$$z=\frac{\pm\frac12\sqrt{x^2+y^2}\mp\frac{y}{\sqrt{x^2+y^2}}}{\pm\frac12\sqrt{x^2+y^2}\pm\frac{y}{\sqrt{x^2+y^2}}}=\frac{x^2+y^2-2y}{x^2+y^2+2y}$$