Let $H$ be a cocommutative hopf algebra. Let $M$ be the category of $H$bimodules. Does the category $M$ form a braided monoidal category with tensor product $\otimes_{H}$ ?

3$\begingroup$ Welcome to MO ! I think not, why should it ? The tensor product doesn't even involve the Hopf structure. $\endgroup$– AdrienMar 25 at 19:29
The answer is no in general.
Here is a counter example. Let us work over a ground field $k$, and let $ H = \oplus_n k$ be the direct sum of $n$ copies of $k$, with $n \geq 2$. This is a commutative, cocommutative Hopf algebra. Of course, as others have commented, the monoidal category $({}_H Mod_H, \otimes_H)$ only depends on the algebra structure of $H$.
The category of $H$$H$bimdoules with $\otimes_H$ monoidal structure has a nice interpretation. An $H$$H$bimodule $M$ can be thought of as an $n \times n$matrix of vector spaces. The $(i,j)$th entry of this matrix is obtained by multiplying $M$ by the $i$th minimal idempotent on one side and the $j$th minimal idempotent on the other side.
Under this identification the monoidal structure $\otimes_H$ is easy to describe: it is "matrix multiplication", but where you replace the usual addition and multiplication operations with direct sum and tensor product of vector spaces.
However, now we see the problem emerge. For $n \geq 2$ matrix multiplication is not commutative, and it is impossible to equip $({}_H Mod_H, \otimes_H)$ with a braiding.
For a concrete example we can set
$$ M = \begin{bmatrix}0 & V\\0 & 0\end{bmatrix}$$
$$ N = \begin{bmatrix}0 & 0\\0 & W\end{bmatrix}$$
Then $N \otimes_H M = 0$, while
$$M \otimes_H N = \begin{bmatrix}0 & V \otimes_k W \\0 & 0\end{bmatrix}$$
There can be no isomorphism between $N \otimes_H M$ and $M \otimes_H N$.