Computing: $\lim_{x\to\infty}\frac{\sqrt{1-\cos^2\frac{1}{x}}\left(3^\frac{1}{x}-5^\frac{-1}{x}\right)}{\log_2(1+x^{-2}+x^{-3})}$

Your idea is very good; the limit you get is for $t\to0^+$, so $\sqrt{1-\cos^2t}=\sin t$ and you get $$ \lim_{t\to0^+}\frac{\sin t(3^t-5^{-t})}{\log_2(1+t^2+t^3)}= \lim_{t\to0^+}\frac{\sin t}{t}\frac{3^t-5^{-t}}{t}\frac{t^2\log 2}{\log(1+t^2+t^3)} $$ (where “log” denotes the natural logarithm) and you can compute separately the limit of the three factors. The first is known to be $1$. Then $$ \lim_{t\to0^+}\frac{3^t-5^{-t}}{t}=\log 3+\log 5 $$ because it's the derivative at $0$ of $f(t)=3^t-5^{-t}$. Alternatively, write it as $$ \lim_{t\to0^+}\left(\frac{3^t-1}{t}+\frac{5^t-1}{t}\frac{1}{5^t}\right) $$ and use the fundamental limits (which is basically what you did).

For the last one, apply l’Hôpital (or Taylor): $$ \lim_{t\to0^+}\frac{2t\log2}{\dfrac{2t+3t^2}{1+t^2+t^3}}= \lim_{t\to0^+}\frac{2(1+t^2+t^3)\log2}{2+3t}=\log2 $$ So finally you get $(\log3+\log5)\log2=(\log 15)(\log 2)$