as a sort of meta answer: there are more left adjoints than right adjoint (this is of course nonsense!!). More to the point, quite often we are interested in finding left adjoints to functors (in order to create free objects (which include all of the uses you mentioned)). Also quite often, the functors that typically interest us do not have right adjoints at all. This is related to the fact of life that mathematics is not self-dual. For whatever reason, many of the naturally occurring functors we meet tend to have left adjoint but often they lack right adjoints.
So, while category theory is self-dual, practiced mathematics is not. A similar phenomenon is the common concept of a cofibrantly generated model category, while there are very very few examples of naturally occurring categories that support the dual notion of a fibrantly generated model category.
Notice that this lack of duality in practiced mathematics starts very early on. The concept of cartesian product of sets is much more important than the dual notion of disjoint union. And while noticing this, it is the cartesian product of set that is at the heart of the definition of category. So the bias towards one concept over its dual is very deeply rooted.
The left adjoint $L_c$ to evaluation-at-$c$ is very simple; left adjoints preserve colimits and every set is a coproduct of $1$ with itself. Consequently,
$$ L_c(X) = X \cdot L_c(1)$$
where $\cdot$ means to take the $X$-fold coproduct of $L_c(1)$ with itself. Finally,
$$ \mathbf{PSh}(\mathcal{C})(L_c(1), F) \cong \mathbf{Set}(1, F(c)) \cong F(c) $$
therefore, $L_c(1)$ is the functor $\mathcal{C}(-, c)$ represented by $c$. That is,
$$ L_c(X) = X \cdot \mathcal{C}(-, c) $$
The right adjoint $R_c$ is even simpler:
$$ R_c(X)(d) \cong \mathbf{PSh}(\mathcal{C})(\mathcal{C}(-, d), R_c(X)) \cong \mathbf{Set}(\mathcal{C}(c, d), X) \cong X^{\mathcal{C}(c,d)}$$
That is,
$$ R_c(X) \cong X^{\mathcal{C}(c, -)}$$
(thanks to Andreas Blass for reminding me of the argument)
The evaluation-at-c functor $F \to F(c)$ is, incidentally, given by the functor
$$ \mathbf{Set}^{\mathcal{C}^\circ} \to \mathbf{Set}^1 $$
induced by the inclusion $1 \to \mathcal{C}^{\circ}$ that identifies $c$, so both adjoints are special cases of the fact this functor has adjoints. I don't remember how this works out off the top of my head, but it should probably be easier to find a reference for.
Best Answer
In general, the evaluation functors have adjoints whenever $D$ is small and $C$ is complete/cocomplete (for right/left adjoints), the adjoints are then precisely Kan extensions along the functor $1\to D$ picking out the object being evaluated at.
In the case of simplicial sets, the right adjoint to evaluation at the $0$-simplex is the 'codiscrete simplicial set' functor that sends a set $S$ to the simplicial set having points of $S$ as vertices, and exactly $1$ $n$-simplex for every $(n+1)$-tuple of points.