Calculate $\lim_{x\to{0^+}}\frac{\log_{\sin{x}}{\cos{x}}}{\log_{\sin{\frac{x}{2}}}\cos{\frac{x}{2}}}$

You are correct. This is another way to proceed: if $x\to 0^+$ then by using Taylor expansions we obatin $$f(x):=\log_{\sin(x)}{(\cos (x))}=\frac{\log(\cos(x))}{\log(\sin(x))}=\frac{\log(1-\frac{x^2}{2}+o(x^2))}{\log (x+o(x^2))}\sim-\frac{x^2}{2\log(x)},$$ and it follows that for any $a>0$, $$\lim_{x\to{0^+}}\frac{\log_{\sin(x)}{\cos(x)}}{\log_{\sin(ax)}\cos(ax)}=\lim_{x\to{0^+}}\frac{f(x)}{f(ax)}=\lim_{x\to{0^+}}\frac{-\frac{x^2}{2\log(x)}}{-\frac{(ax)^2}{2\log(ax)}}=\lim_{x\to{0^+}}\frac{\log(x)+\log(a)}{a^2\log(x)}=\frac{1}{a^2}.$$ In your case $a=1/2$ and the limit is $4$.