E1 reactions occur in two steps. First, the leaving group is removed, yielding a carbocation. Second, a weak base removes a proton from the carbon adjacent to the carbocation carbon. Thus, to favor E1, a protic solvent is desired in order to stabilize the carbocation. Weak bases favor an E1 mechanism.

E2 reaction occur in only one step. The strong base removes the beta-hydrogen, while the leaving group simultaneously leaves. The double bond forms simultaneously. 

Elimination reactions involve the use of bases, which remove hydrogen atoms. The leaving group, which is bound via a sigma bond is removed, along with a hydrogen (thus two sigma bonds are broken). The result is a double bond, which consists of one sigma bond and one pi bond. 

E1 reactions occur in two steps, forming a carbocation intermediate, which is most stable if there is a protic solvent present. Furthermore, the use of a weak base favors E1.

E2 reactions occur in one step; thus no carbocation intermediate is formed, and an aprotic solvent is favored. E2 reactions are favored by strong bases and higher temperatures.

These conditions are perfect for an  reaction. We see a secondary halide so first instinct might be to say that an  reaction would ensue. However, anytime we have a secondary halide reacting with a strong, bulky base, the laws of sterics dictates that an  reaction will ensue.

The given conditions are perfect for an  reaction. Here we have a primary haloalkane reacting with a big, bulky base. Big, bulky bases such as , and alkene products are classic indicators that the reaction was an  reaction.

Treatment with strong base indicates E2 mechanism. The halide has two beta-carbons (II and III), and potassium tert-butoxide is a bulky base. The favored product is the Hofmann product (less substituted alkene). Therefore III is the hydrogen that's removed. 

E2 reaction mechanism requires antiperiplanar orientation between the leaving group and the hydrogen. Given the stereochemistry of the methyl group, forming an alkene bond (to give the more substituted product) is impossible.

This is an acid-catalyzed dehydration of a tertiary alcohol. The first step is protonation of the alcohol oxygen which will create a good leaving group (), and subsequent loss of water from the molecule. The molecule will then contain a carbocation (which will not move due to the positive being on the most substituted carbon atom in its immediate vicinity), and the conjugate base (water in this case) will remove a hydrogen from a carbon next to the carbocation. The base will preferentially remove a hydrogen from the one that will produce the more substituted alkene, as this is generally the more stable product. This is known as Zaitsev's rule for the formation of alkenes.