Question: can anyone tell me, please, two ultrafilters such that $\mathcal{U}\otimes\mathcal{V}\neq\mathcal{V}\otimes\mathcal{U}$ and others two such that $\mathcal{U}\oplus\mathcal{V}\neq\mathcal{V}\oplus\mathcal{U}$?. Thanks

Answer: 
For $\mathcal{U}\oplus\mathcal{V}$:

Let $X=\{x\_1,x\_2,\dots\}$ and $Y=\{y\_1,y\_2,\dots\}$ be two infinite subsets of $\mathbb{N}$ with this property: $A=\{x\_i+y\_j;\,i<j\}$ and $B=\{x\_i+y\_j;\,i>j\}$ satisfy $A\cap B=\emptyset$. For example, $x\_i=2^{2i}$ and $y\_i=2^{2i+1}$ do satisfy this.

Let now $U=\{C\subset\mathbb{N};\,X\backslash C\text{ is finite}\}$. $U$ is a filter, so let us enlarge it to an ultrafilter $\mathcal{U}$. Similarly we construct $V$ and $\mathcal{V}$ (using $Y$ in place of $X$).

We now have $A\in\mathcal{U}\oplus\mathcal{V}$ (since for every $x\_i$, the set $A-x\_i$ contains as a subset $\{y\_{i+1},y\_{i+2},\dots\}$, so $A-x\_i\in V\subset\mathcal{V}$). Similary, $B\in\mathcal{V}\oplus\mathcal{U}$. Since $A\cap B=\emptyset$, we have $\mathcal{U}\oplus\mathcal{V}\neq \mathcal{V}\oplus\mathcal{U}$.
For $\mathcal U\otimes \mathcal V$, let $\mathcal U=(0)$ (principal), $\mathcal V$ -- any other. Then $A=\mathbf N\times \lbrace 0\rbrace$ is not in $\mathcal U\otimes \mathcal V$, (since all vertical sections of $A$ are $\lbrace 0\rbrace\notin \mathcal V$), but is in $\mathcal V\otimes \mathcal U$ (since all vertical sections of $A$ contain $0$).