Proving $k^{m+l} = k^m k^l$ by constructing a bijective function F : $ ^MK \times ^LK \to ^{L\bigcup M}K $ 
*

*For cardinals k which is cardinal of K and l which is cardinal of L and m which is cardinal of M.
W.T.S [ $ k^{m+l} $ = $ k^m k^l $] by constructing a bijective function F : $ ^MK \times ^LK \to ^{L\bigcup M}K $


I construct a function F 
$ F(f,g)(x) = f(x) , if~x \in L, F(f,g)(x) = g(x), if~ x\in M $
and $ F(f,g)(x) = f(x) ~if~\in M \bigcap L $ 
Is it right?


*For cardinal $ 2\leq k \leq l $ where $l$ is infinite, W.T.S [$ 2^l = k^l = l^l $] 


my answer is 
$ 2^l \leq 2^{l \cdot l}\leq2^l \cdot 2^l \leq k^l \cdot 2^l=k^l \leq k^l \cdot k^l \leq l^l \cdot k^l \leq l^l $
How can I do it in reverse direction for using Cantor-Bernstein Theorem? 


*For an infinite set B and a subset A of B such that $card(A)< card(B) $, 


W.T.S [$ Card(B-A) = Card (B) $]
I have totally no idea of this problem. 
 A: The function $F$ that you constructed for the first problem is almost right, but you interchanged $M$ and $L$. You want
$$[F(f,g)](x)=\begin{cases}
f(x),&\text{if }x\in M\\
g(x),&\text{if }x\in L\;.
\end{cases}$$
Of course you do still have to verify that $F:{^MK}\times{^LK}\to{^{M\cup L}K}$ is a bijection.
For the second question, how do you know that $2^{\ell\cdot\ell}\le 2^\ell\cdot2^\ell$? 
To show that $2^\ell\le k^\ell\le\ell^\ell$, show that if $a,b$, and $c$ are any cardinals such that $a\le b$, then $a^c\le b^c$; use an injection from a set $A$ of cardinality $a$ into a set $B$ of cardinality $b$ to get an injection from ${^CA}$ to ${^CB}$, where $C$ is a set of cardinality $c$. For the other direction use the fact that $\ell\le 2^\ell$, so, by the result just discussed, $\ell^\ell\le(2^\ell)^\ell$, and prove that $(2^\ell)^\ell=2^{\ell\cdot\ell}=2^\ell$. The proof that $(2^\ell)^\ell=2^{\ell\cdot\ell}$ is somewhat similar to the first question.
In this answer I proved the special case of the third question in which $A$ is countable and $B$ is uncountable; try to adapt it to cover the general situation.
