I am posting this answer to make solutions more clear.
$$ \mathsf{SIII}=_\beta (λxyz.(xz)(yz)) \mathsf{III} =_\beta (\mathsf{II})(\mathsf{II}) $$
Remember that $\mathsf{I}=\lambda x.x$ is the operator that does nothing. This means that for every lambda expression $M$, you have $\mathsf{I}M =_\beta M$.
This implies that:
$$\mathsf{II} =_\beta (\lambda x.x)(\lambda x.x)=\lambda x.x = \mathsf{I}$$
And so we have
$$ (\mathsf{II})(\mathsf{II}) =_\beta (\mathsf{I})(\mathsf{I}) =_\beta = \mathsf{II} =_\beta \mathsf{I}$$
The second one is easier
$$ \operatorname{twice} \operatorname{twice} f x = \operatorname{twice} (\operatorname{twice} f) x = \operatorname{twice} f (\operatorname{twice} f x) = \operatorname{twice} f f(f(x) = f(f(f(f(x)))) $$