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).

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).

Later edit: many people have complained about the limited usefulness of the adjoint functor theorems in that in many cases there is a direct, and thus much more enlightening, construction. But there are situations where such a direct construction is not available. One that I came across recently is when studying P. Johnstone's book Stone spaces, more precisely chapter III and the section on Manes' theorem about the monadicity of the category of compact Hausdorff spaces. In the sequel, P. Johnstone proves another result due to Manes, the fact that category of algebras (in the sense of universal algebra) in the category of compact Hausdorff spaces is also monadic. He remarks that one has to use the GAFT (and Beck's monadicity theorem) in this case, because there is no easy direct description of the left adjoint. Later in the book (somewhere, I am quoting from memory and do not have the book by me), he argues why there is no simple recipe for the left adjoint.