$X,Y$ independent then $X+Y$, $X-Y$ independent as well?

Not always. For example, let, $X$ be the number obtained on rolling a die, and $Y$ the number obtained on rolling another die. Then clearly $X$ and $Y$ are independent. However, for example, $$P(X+Y{=}\,6\,\hbox{and}\,X-Y{=}\,1)=0\ne P(X+Y{=}\,6)P(X-Y{=}\,1)\ ,$$ so $X+Y$ and $X-Y$ are not independent.

$\newcommand{\cov}{\operatorname{cov}}\newcommand{\var}{\operatorname{var}}$ $$ \cov(X+Y,X-Y) = \var X -\var Y \\[8pt] \begin{cases} =0 & \text{if }\var X=\var Y, \\ \ne 0 & \text{if }\var X\ne\var Y. \end{cases} $$ In the cases where this $\ne0$, the random variables $X+Y$ and $X-Y$ cannot be independent.

In some of the cases where this $=0$, they are independent, but as "David"'s answer has shown, not in all.