There are two parts to this answer.

1. First, a functor must be continuous (cocontinuous) to have a left (right) adjoint. Most of the times, it is easy to check that a functor does *not* preserve (co)limits and thus it cannot have a a left (right) adjoint.

2. (co)continuity is not enough to actually prove that a functor has the required adjoint, but it is *almost* good enough. Let me elaborate on this. If you have a functor $F:P\to Q$ between complete partial orders (and thus cocomplete) then it is an easy exercise to construct a left adjoint by taking a $\sup$ of an appropriate subset. This can be generalized in a straightforward way to any functor by taking an appropriate (co)limit. The bad news is that this (co)limit is in general over a large category so it may not exist. This is where the so-called solution-set conditions come in; they are way to trim down this large category to a small one.

As many people already said there are various variations of this type of conditions, from the more general but also very cumbersome to check solution-set condition to easier conditions which combine some form of well-poweredness (each object has only a set of subobjects -- or quotients, whatever the case may be) with the existence of a small separating (or generating) set. One that guarantees the existence of a right adjoint and that sticks out particularly in my memory is the existence of a small dense subcategory -- check chapter V of Kelly's book on enriched category for the precise details. It is particularly memorable, because many categories come with god-given small dense categories like presheaf categories (courtesy of Yoneda) and sheaf categories (because dense composed with left adjoint is dense).