Proof (1) is not precise but the idea is correct. For (2), I don't see any proof.
Let $f$ strictly convex, and suppose that there are two global minimums at $x_0$ and $x_1$ (where $x_0<x_1$). Let $\lambda \in (0,1)$. Then $$f(x_0)\leq f\big(\lambda x_0+(1-\lambda )x_1\big)< \lambda f(x_0)+(1-\lambda )f(x_1)$$
$$\underset{f(x_1)\leq f(x_0)}{\leq} \lambda f(x_0)+(1-\lambda )f(x_0)=f(x_0),$$
which is a contradiction.
Simplifying gives: $$\frac{\sin x}{\cos y}+\frac {\cos x}{\sin y}+\frac{\sin y}{\cos x}+\frac{\cos y}{\sin x}=\frac{\sin^2 x+\cos^2y}{\sin x\cos y}+\frac {\cos^2 x+\sin^2y}{\sin y\cos x}=\frac{\sin(x+y)\sin(x-y)+1}{\sin x\cos y}+\frac {1-\sin(x+y)\sin(x-y)}{\sin y\cos x}=\frac {8}{3\sin2x\sin 2y}+\frac{2\sin (x-y)}{3}\left(\frac{1}{\sin x\cos y}-\frac{1}{\sin y\cos x}\right)=\frac {8}{3\sin2x\sin 2y}-\frac 83\frac{\sin^2(x-y)}{\sin2x\sin 2y}=\frac{8\cos^2(x-y)}{3\sin 2x\sin2y}=\frac{16\cos^2(x-y)}{3(\cos 2(x-y)-\cos 2(x+y))}=\frac{16}{3(2-10/9\sec^2(x-y))}$$
The denominator is maximum when $\sec^2(x-y)$ is minimum, which happens when $x=y$, in which case the quantity has value $6$. So the required global minimum should be $6$.
Edit: The edit shows that the given expression (let's call it $f(x,y)$ for brevity) can't take any value smaller than $6$. To see that, let's first note (Refer Note) that $\color{blue}{ 2-\frac {10}9\sec^2(x-y)\gt 0 \text{ for all } (x,y)\in (0,\frac \pi 2)\times (0,\frac \pi 2)}$.
Suppose on the contrary that there exist some $a$ and $b$ in $(0,\frac \pi 2)$ such that $f(a,b)<6$. It follows that $$\frac{16}{3(2-10/9\sec^2(x-y))}\lt 6\implies 8<18-10\sec^2(x-y)\implies \sec^2(x-y)\lt 1 $$ and this is a contradiction. So the assumption that $f$ takes any value less than $6$ is wrong. Hence global minimum value of $f$ under the given conditions is $6$.
Note: If $\sec^2(x-y)\ge \frac{18}{10}$ then $\cos^2(x-y)\le \frac {10}{18}\implies 0\lt \cos (x-y)\le \sqrt {\frac{10}{18}}$ (because $x-y \in (-\frac \pi 2, \frac \pi 2)$ so $\cos $ is +ve) so let's consider two cases: 1) $0\le x-y)\lt \frac \pi 2 \text{ and } 2) -\frac \pi 2\lt x-y \lt 0$.
Case 1: it follows that $x-y\ge \arccos\sqrt {\frac{10}{18}}$ (noting that $\cos$ is decreasing on $[0,\pi/2))$. Given that $x+y=\arcsin \frac 23$, it follows that $y\le 0$ which is not possible.
Case 2: it follows that $x-y\le -\arccos\sqrt {\frac{10}{18}}$ (noting that $\cos$ is increasing on $(-\frac \pi 2, 0))$. Then proceeding as in case 1) results in a contradiction.
This establishes the blue colored part.
Best Answer
It follows from $$ \alpha^2 + \beta^2 \le (|\alpha|+|\beta|)^2<1 $$ that $$ f''(x) = 1-\alpha^2\cos \alpha x - \beta^2 \sin \beta x \ge 1-\alpha^2 - \beta^2 > 0 \, , $$ i.e. $f$ is strictly convex. The existence of a global minimum follows from the fact that $$ \lim_{x \to -\infty} f(x) = \lim_{x \to +\infty} f(x) =+\infty \, , $$ and the strict convexity ensures that there is no other (local or global) minimum.