For (1), you need to verify that for each continuous and bounded $h$, the sequence $Eh(f(X_n,Y_n)))$ converges to $Eh(f(X,Y))$. But the composition $g(x,y)=h(f(x,y))$ is continuous and bounded if $h$ is c&b and if $f$ is continuous.
So the weak convergence of $(X_n,Y_n)$ implies the weak convergence of $f(X_n,Y_n)$.
Similarly for (2), except now you need to check that for each continuous and bounded $h(x,y,z)$ that $Eh(X_n,Y_n,f(X_n,Y_n))$ converges to $Eh(X,Y,f(X,Y))$. In this case you mentioned no hypothesis on $f$, but continuity will allow the above argument to go through.
As follows. For given bounded continuous $h$ define the function $g$ of two variables by $g(x,y)=h(x,y,f(x,y))$. Assuming $f$ is continuous, so is $g$; clearly $g$ is also bounded. By assumption $(X_n,Y_n)$ converge in distribution to $(X,Y)$, so $ Eg(X_n,Y_n)\to Eg(X,Y)$. But this means $ Eh(X_n,Y_n,f(X_n,Y_n)) \to E h(X,Y,f(X,Y))$, by the way $g$ is defined in terms of $h$.
Since this holds for all continuous and bounded $h$ we see that $(X_n,Y_n,f(X_n,Y_n))$ converges in distribution to $(X,Y,f(X,Y))$.
Best Answer
Yes, because if we look at the joint characteristic function:
$$Ee^{i(uX_n + vY_n)} = Ee^{iuX_n}Ee^{ivY_n} \to Ee^{iuX}Ee^{ivY} = Ee^{i(uX + vY)}$$