Semisimple modules and the radical I don't need a proof, but  can someone tell me whether it is true that for all $A$-modules $V$ we have that $V/\text{rad}V $ is semisimple, where we define $\text{rad} V$ as the intersection of all maximal submodules of $V$?
 A: For non-finite-dimensional algebras certainly not. 
Take $A=k[X]$ and $V=k[X],$ the regular module. Then $\mathrm{rad} V=0,$ basically since $k[X]$ is a PID with infinitely many maximal ideals*, but $V$ is not semisimple (because, for example, all simple $A$-modules are easily seen to be finitely dimensional. However, $A$ is not, and if $A$ was semisimple, it would be a sum of finitely many simples).
(Explanation of *: in a PID, the claim on maximal ideals is equivalent to the statement that there are infinitely many irreducible elements. To check this for $k[X]$, simply invoke the classical Euclid's argument for  infinitude of primes, i.e. "multiply together all irreducible elements and add 1 to show that there is at least one not on the list").
A: In the unlikely case by semisimple you mean "has zero radical" (this is sometimes called "Jacobson semisimple" ) then yes, it's always true that $\mathrm{rad}(V/\mathrm{rad}(V))=\{0\}$.
But if you mean "is a direct sum of simple modules" then (as PavelC has already pointed out)  there are many counterexamples. The polynomial ring is nice, but I'll try to give another one. If $F$ is any field, then $\prod_{i=1}^\infty F=A=V$ has radical zero, but $V_A$ is not semisimple. It's not semisimple because $\oplus_{i=1}^\infty F\subseteq V$ does not have a direct complement in $V$. 
