A quick Google search found this paper, where it is stated that the answer to the first question is yes in the compact case (due to Tognoli): this result is called the Nash-Tognoli theorem. In general, the answer is no: a real affine variety has finite-rank homology groups, and it's easy to construct non-compact manifolds for which this is false (e.g. a surface of infinite genus). In fact, apparently there is a bound due to Milnor for the sum of the Betti numbers of a real variety.
The answer to the second question is certainly not: just take two elliptic curves with slightly different $j$-invariants.
A) A smooth algebraic subvariety $X\subset \mathbb C^n$ is locally on X given by the set $V(f_1,\cdots,f_k)\subset \mathbb C^n$ of common zeros of a list $f_1,\cdots,f_k$ of polynomials with Jacobian matrix of rank $k$.
The implicit function theorem then shows that if we consider these polynomials as holomorphic functions, the variety $X$ is a holomorphic submanifold of $\mathbb C^n$.
Beware that in general it is impossible to find polynomials $f_1,\cdots,f_k$ such that $X=V(f_1,\cdots,f_k)$ and which satisfy the Jacobian condition at every $x\in X$ : the polynomials $f_i$ a priori depend on the choice of $x\in X$.
(If however one can find such polynomials working everywhere independently of $x\in X$, then $X$ is said to be a complete intersection in $\mathbb C^n$.)
B1) The simplest complex holomorphic manifold $X$ with no algebraic structure is $X=\mathbb C\setminus \mathbb Z$.
The precise statement is that there does not exist an algebraic variety $Y$ whose analytification is (isomorphic to) $X$ i.e. $Y^{an}=X$is impossible. We then say that $X$ is not algebraizable
The reason why $\mathbb C\setminus \mathbb Z$ is not algebraizable is that any complex non compact smooth algebraic variety of dimension $1$ is obtained by deleting finitely many points from a compact one-dimensional smooth variety.
So there is a purely topological obstruction to $\mathbb C\setminus \mathbb Z$ being algebraizable.
B2) Every one-dimensional compact complex manifold is algebraizable: this is celebrated theorem of Riemann.
B3) There exist compact complex tori $X$ of any dimension $n\geq 2$ which are not algebraizable.
A complex torus is a manifold of the form $X=\mathbb C^n/\Gamma$ where $\Gamma =\oplus _{j=1}^{2n}\mathbb Z\cdot v_j$ is the lattice obtained from some basis $(v_j)_{j=1\cdots 2n}$ of the real vector space underlying $\mathbb C^n$.
The subtle relations between the $v_j$'s dictating whether $X$ is algebraizable or not were discovered by Riemann (him again!) and are referred to as Riemann bilinear relations.
Edit (September 2, 2016): the technical condition for algebraizability of a torus.
The Riemann criterion for algebraicity of the torus $\mathbb C^n/\Gamma$ is that there exist a hermitian form $H:\mathbb C^n\times \mathbb C^n\to \mathbb C$ whose imaginary part is integral on the lattice:$$(Im H)(\Gamma\times \Gamma)\subset \mathbb Z$$
For example every complex torus of dimension $1$ is obtained by dividing out $\mathbb C$ by a lattice of the form $\Gamma =\mathbb Z \oplus \mathbb Z\tau$ where $\tau =a+ib$ with $b\gt0$.
Such a torus is algebraizable as witnessed by the hermitian form $H(z,w)=\frac {z\overline w}{b}$.
This is of course in line with the result mentioned in 2.
Bibliography
A nice introduction to these ideas is Shafarevich's Basic Algebraic Geometry 2: Chapter 8, page 153.
Best Answer
You are absolutely right that a complex variety with its Zariski topology is not a complex manifold, nor even a Hausdorff topological space (unless it has dimension zero).
However there is a completely canonical way of associating to a complex algebraic variety $X$ a complex analytic variety $X^{an}$.
More precisely that association Algvar $\to$ Anvar is a functor.
This functor has been studied in detail by Serre in a ground-breaking article published in 1956 and universally known by its amusing acronym GAGA.
A typical result in the article (Proposition 6, page12) is that $X$ is complete iff $X^{an}$ is compact: a highly non-trival result relying on a theorem of Chow.
In this set-up the result you are asking about can be stated as follows:
An algebraic complex variety $X$ is regular (=smooth) if and only if the associated analytic variety $X^{an}$ is a complex manifold.
Edit
Here is an English translation of GAGA.