Hofmann rule and deviation from E2 mechanism
The difference between the different eliminations like E1, E2 or Hoffmann elimination is the proper choice of the base and proper solvent medium.

Let's take your example i.e. 2-bromo-2-methylbutane, and choose different bases and reaction media for observing differences between E1, E2 and Hoffmann elimination.



Why not E1?

First let's consider the reaction as:

$$\ce{2-bromo-2-methylbutane + EtO^- (in EtOH) -&gt; }$$

Clearly, this reaction is conducted in a highly polar solvent where the carbocation - formed when $\ce{Br^-}$ leaves - gets stabilised. The base then abstracts the hydrogen of $\ce{C3}$ to form 2-methylbut-2-ene by an E1 mechanism.

If you use the same substrate but this time in a highly non-polar solvent, the possibility of E1 decreases as carbocation formation won't get extra stability. But as the base is not sterically hindered, it can still abstract hydrogen to form the Saytzeff product, but there is a case where it can also form the Hoffmann product.



Why not E2?

The E2 mechanism follows an "anti-periplanar orientation of hydrogen and leaving group". Suppose the reactant's stereochemistry is such that there is no anti-$\ce{H}$ on the more substituted adjacent carbon for forming the Saytzeff product, but there is an anti-$\ce{H}$ on a less substituted adjacent carbon, which will lead to the Hoffmann product. In that case, the major product can be the less substituted alkene due to stereospecificity of E2 reaction.



Why E1cb?

Now consider your above given case. Here, the base $\ce{t-BuO^-K+}$ is sterically hindered and is a very strong base. So, prior to any carbocation formation, this base will abstract hydrogen from the less hindered position, and form a carbanion. Also, a less substituted carbanion is more stable than a more substituted one (due to a smaller +I effect). Both these effects lead to formation of a less substituted alkene via an E1cb mechanism.



So, depending upon which product is needed as a major one and the proper stereochemistry of reactant, you should choose whether to go with an E1, E2, or E1cb mechanism by choosing the proper base (bulky or less hindered) and proper solvent (polar or non-polar).