We'll deal with the additional question first because it's easily set down: $V(\mathfrak{a})$ is irreducible iff $\sqrt{\mathfrak{a}}$ is prime, and this is true in complete generality. But there are plenty of examples where $V(\mathfrak{a})$ is irreducible yet $\mathfrak{a}$ is not prime: take $(x^2)\subset k[x]$, for instance.
The first statement is a little more involved. It is not true that $V(f)$ irreducible with $f$ irreducible implies $(f)$ prime. Consider $V(xy-z^2)\subset\Bbb A^3$ with coordinate algebra $k[x,y,z]/(xy-z^2)$ and take $f=x$. Then $V(x)\subset V$ is just the $y$-axis, which is clearly irreducible, and $(x)$ is not prime because $(k[x,y,z]/(xy-z^2))/(x)\cong k[x,y,z]/(x,xy-z^2)\cong k[y,z]/(z^2)$ is not a domain. All that's left is to verify that $x$ is irreducible.
To show that $x$ is irreducible, consider the norm map $N:k[x,y,z]/(xy-z^2)\to k[x,y]$ which sends $p(x,y)+q(x,y)\cdot z$ to $(p+qz)(p-qz)=p^2-xyq^2$. $N$ is multiplicative, and I claim that $N(a)$ is a unit iff $a$ is a unit. One direction is easy: if $ab=1$, then $1=N(1)=N(ab)=N(a)N(b)$. On the other hand, suppose $q\neq 0$. Then for $p^2-xyq^2=u$ to be true, $\deg q=\deg p-1$ and looking at highest homogeneous parts, we must have $(p)_{\deg p}^2=xy(q)_{\deg q}^2$, where $(-)_d$ denotes the degree $d$ part of that polynomial. But this cannot be true: if $x\mid (p)_{\deg p}^2$, then $x^2\mid (p)_{\deg p}^2$ and similarly with $y$, and we reach a contraction by descent. So $q=0$ and therefore $p=\sqrt{u}$.
Therefore to prove that $x$ is irreducible it suffices to show that there is no element $p(x,y)+q(x,y)z$ so that $p^2-xyq^2=ux$. Evaluating both sides at $y=0$, we find that $p(x,0)^2=ux$ which is impossible by looking at the top-degree term of $p(x,0)$. So $x$ is irreducible and we're done.
Best Answer
There is a corresponding statement for varieties, and it's exactly what you intuit.
To see why this is the case, recall we have an (inclusion reversing) correspondence between radical ideals and subvarieties of $V$. So asking if
$$\mathcal{V}(\mathfrak{a}) \supseteq \{ x \}$$
is asking if the variety of $\mathfrak{a}$ contains a point $x$ (that is, a minimal nonempty subvariety $\{ x \}$). By the inclusion reversing correspondence, this is the same as asking if
$$\mathfrak{a} \subseteq \mathcal{I}(\{x\})$$
that is, if $\mathfrak{a}$ is contained in the ideal associated to a point $x$ (that is, a maximal proper ideal).
Of course, we know that every proper ideal is contained in some maximal proper ideal (by Zorn's Lemma), and now by the nullstellensatz (since $k$ is algebraically closed) we know that maximal ideals of $k[V]$ are all of the form $(x_1 - a_1, \ldots, x_n - a_n)$ for some point $(a_1, \ldots, a_n) \in k^n$.
If you're not sure why we can apply the nullstellensatz here, it's because ideals in $k[V] = k[x_1, \ldots, x_n] / I$ are exactly the ideals of $k[x_1, \ldots, x_n]$ containing $I$, so we get the claim by the "ordinary" nullstellensatz followed by the correspondence principle.
So, to put all the pieces together, how would you actually write up a proof of this fact? You should try it yourself (everything you need is in this answer) and check against the version I've written up under the fold:
I hope this helps ^_^