I'm looking for an example in the literature where $\mbox{Pic}^0(X)$, $\mbox{Pic}(X)$, and $NS(X)$ of a projective surface $X$ over a field are calculated. I want them for an example I'm trying to work out, so ideally $X$ would be relatively simple, perhaps a cubic hypersurface in $\mathbb{P}^3$, or something along those lines. I know it's out there, but googling and browsing arXiv and MathSciNet haven't quite panned out.

Manin's book "Cubic forms" contains the calculations of these groups when $X$ is a smooth projective cubic surface. In particular, $\operatorname{Pic}^0(X)=\{0\}$ and $\operatorname{Pic}(X)=\operatorname{NS}(X)$ is a free commutative group of rank 7. Another class of examples is provided by products $X=E \times E'$ of two elliptic curves. In particular, if $E=E'$ has no complex multiplication then $NS(E\times E)$ is a free commutative group of rank 3 generated by the classes of $E \times \{0\}$, $\{0\}\times E$ and the diagonal while $\operatorname{Pic}^0(E\times E)=E \times E$. See Mumford's Abelian Varieties. 


For surfaces in $\mathbb{P}^3$ of degree at most 3 the calculation of $Pic(X)$ is relatively easy: In this case $X$ is rational, hence $NS(X)$ modulo torsion equals $H^2(X,\mathbb{Z})$. (If you work over the complex numbers you might also apply Lefschetz (1,1)Theorem.) Starting from degree 4 the calculation of $NS(X)$ is much more involved. The difficulty depends on how you present $X$. In case you give $X$ just by an equation it is not so easy to calculate $NS(X)$, at least if you work in characteristic 0. See e.g., http://pjm.math.berkeley.edu/ant/2007/11/p01.xhtml where an example is given of a quartic surfaces with $\mathrm{rank} Pic(X)=\mathrm{rank} NS(X)=1$. Over a finite fields (or over $\overline{\mathbb{F}_p}$) you can find an upper bound for the rank of $NS(X)$ in terms of the zeta function of $X$ (see loc. cit.). If you believe the Tate conjecture then this upper is the actual rank of $NS(X)$. In concrete examples you might try to use this upper bound, try to find sufficiently many curves on $X$ and then use the intersection pairing to prove that the classes of these curves are independent in $NS(X)$. 


In general, if $X$ is a smooth complex projective variety which is simply connected, then we have $\rm{Pic}^0(X)=0$. Indeed we have $H^1(X,\mathbb{Z})=0$, and then Hodge theory implies that $H^1(X,\mathcal{O}_X)=0$. The exponential sheaf sequence http://en.wikipedia.org/wiki/Exponential_sheaf_sequence then implies that the natural map $\rm{Pic}(X) \to H^2(X, \mathbb{Z})$ is injective. In particular, any hypersuface of dimension greater than $1$ is simply connected (by the Lefschetz hyperplane section theorem), and so $\rm{Pic}^0(X)$ is always trivial in this case. 


There are also some computations for conic bundles by Sansuc in MR0695346 (85d:14014) Sansuc, JeanJacques À propos d'une conjecture arithmétique sur le groupe de Chow d'une surface rationnelle. 


For Reid's list of 95 K3 surfaces Picard lattices have been computed by Belcastro. Her paper can be downloaded from the arXiv at http://arxiv.org/PS_cache/math/pdf/9809/9809008v2.pdf . She has also made her thesis available, which can be downloaded at http://www.toroidalsnark.net/sm_thesis.pdf . 

