If $$\lim_{x\rightarrow0}[\lim_{y\rightarrow0}\frac{1-\cos(x)}{\sin(x)+\sin(y)}]=a$$
$$\lim_{y\rightarrow0}[\lim_{x\rightarrow0}\frac{1-\cos(x)}{\sin(x)+\sin(y)}]=b$$
Then the values of $a$ and $b$ are
(a) No such $a$ and $b$ exists.
(b)$a=0$ but no such $b$ exists.
(c)$b=0$ but no such $a$ exists.
(d)$a=0 $ and $b=0$
$$a=\lim_{x\rightarrow0}[\frac{1-\cos(x)}{\sin(x)}]\implies a=\frac{1+\sin(x)}{\cos(x)}=1$$ $\tag{By L'Hopital's rule}$
Similarly,$b=1$
Note that
$$\lim_{x\rightarrow0}\left[\lim_{y\rightarrow0}\frac{1-\cos x}{\sin x+\sin y}\right]=\lim_{x\rightarrow0}\frac{1-\cos x}{\sin x}=\lim_{x\rightarrow0}\frac{1-\cos x}{x^2}\frac{x^2}{\sin x}=0$$
$$\lim_{y\rightarrow0}\left[\lim_{x\rightarrow0}\frac{1-\cos x}{\sin x+\sin y}\right]=\lim_{y\rightarrow0}\frac{0}{\sin y}=0$$