First notice that we have a bijective function from $\mathbb{N}$ to $\mathbb{Z}$, So we can reduce our problems to $[0,1]^\mathbb{N}$ and $\mathbb{N}^{[0,1]}$. Further we have that $\mathcal{P}(\mathbb{N})$ is uncountable. Then look at the function
$$
f:\mathcal{P}(\mathbb{N})\to[0,1]^{\mathbb{N}}
$$
by sending $A\in\mathcal{P}(\mathbb{N})$ to the map in $[0,1]^{\mathbb{N}}$ which send $x\in\mathbb{N}$ to $1$ if $x\in A$ and zero otherwise. This gives an injective function, thus we conclude that $[0,1]^\mathbb{N}$ is uncountable.
Then for the other notice you can imbed $\mathbb{N}$ in $[0,1]$ by looking at $1/n$. then you can look at a subset of the set $\mathbb{Z}^{[0,1]}$ which is $\{0,1\}^\mathbb{N}$ and then you can say again with the above argument that it is uncountable.