The answer is still **No**, but for not that obvious reason. To show that, let us start with a positive claim.

**1.** FIrst of all, an operator $T$ is semisimple iff all the factors in the prime expansion of its minimal annihilating polynomial $\mu$ are distinct. Actually, the algebra $K[T]$ is isomorphic to $K[X]/(\mu)$; so, if there are no multiple factors, then this algebra is a direct sum of fields, and each its finitely generated module is semisimple. Otherwise, if $\mu=p^2q$ with $p$ nonconstant, then the annihilator space of $p(T)$ has no complement.

**2.** Now assume that the extension of $K$ generated by all the roots of characteristic polynomials of $T_1$ and $T_2$ is separable. Then they should be diagonalizable over this extension by the reasons of the minimal polynomial. Since they commute, they are simultaneously diagonalizable. Hence their sum and product are also diagonalizable, and their minimal polynomials have no multiple roots (from separability!) and hence no multiple factors. So $T_1+T_2$ and $T_1T_2$ are both semisimple.

**3.** And here is a counterexample for non-separable case. Let $K=F_2(t)$ be the field of rational fractions over $F_2$. Set $T_1=\begin{pmatrix}0&0&1&0\cr 0&0&0&1\cr t&0&0&0\cr 0&t&0&0\end{pmatrix}$ and $T_2=\begin{pmatrix}0&1&0&0\cr t&0&0&0\cr 0&0&0&1\cr 0&0&t&0\end{pmatrix}$; their common minimal polynomial is $X^2-t$, so they are semisimple (this polynomial is irreducible although not separable). But their sum is $S=T_1+T_2=\begin{pmatrix}0&1&1&0\cr t&0&0&1\cr t&0&0&1\cr 0&t&t&0\end{pmatrix}$ with $S^2=0$, and their product is $P=T_1T_2=T_2T_1=\begin{pmatrix}0&0&0&1\cr 0&0&t&0\cr 0&t&0&0\cr t^2&0&0&0\end{pmatrix}$ with the minimal polynomial $(X+t)^2$. Hence both are not semisimple. One may present a direct example of a spaces that cannot be complemented as $\langle e_2+e_3,e_1+te_4\rangle$ in both cases.

In fact, this example was obtained from the action of algebra $K[X,Y]/(X^2-t,Y^2-t)$ on its regular module; $T_1$ and $T_2$ correspond to $X$ and $Y$, respectively.

Yes, because $V$ is the direct sum of the eigenspaces $V_\mu$ under $A$. By commutation, $BV_\mu\subset V_\mu$. Hence we are down to look at the restrictions of $A+B$ to each $V_\mu$, which is nothing but $\mu{\rm id}+B|_{V_\mu}$. The latter is semis-imple, hence $A+B$ is too. – Denis Serre Dec 3 '12 at 16:04