Yes, every algebraic differential form is holomorphic and yes, the differential preserves the algebraic differential forms. If you are interested in projective smooth varieties then every holomorphic differential form is automatically algebraic thanks to Serre's GAGA. This answers (3).

Concerning (1) and (2) I suggest that you consult some standard reference as Hartshorne's Algebraic Geometry.

**Edit**: As pointed out by Mariano in the comments below there are subtle points when one compares Kahler differentials and holomorphic differentials. I have to confess that I have not thought about them when I first posted my answer above.

Algebraic differential $1$-forms
over a Zariski open set $U$ are elements of the module generated by $adb$ with $a$ and $b$
regular functions over $U$ (hence algebraic) by the relations $d(ab) = adb + b da$, $d (a + b) = da + db$ and $d \lambda =0$ for any complex number $\lambda$. Since these are the rules of calculus there is a natural map to the module of holomorphic $1$-forms over $U$. **This map is injective**, since the regular functions on $U$ are not very different from quotients of polynomials.

If instead of considering the ring $B$ of regular functions over $U$ one considers the ring $A$ of holomorphic functions over $U$ then one can still consider its $A$-module of Kahler differentials. If $U$ has sufficiently many holomorphic functions, for instance if $U$ is Stein, then one now has a surjective map to the holomorphic $1$-forms on $U$ which is **no longer injective** as pointed out by Georges Elencwajg in this other MO question.