Answer:
Reaction 1 - elimination
Reaction 2- substitution
Reaction 3- the two combined
Step-by-step explanation:
In the first reaction, a strong base, KOH is used in the reaction. Remember that when a tertiary alkyl halide is the substrate, the use of a strong base leads to the domination of E1 mechanism over SN1 mechanism, hence the product shown in the image attached.
In reaction 2, the secondary alkyl halide must undergo an SN2 substitution since an aprotic solvent (THF) is used. This leads to inversion of configuration as shown in the product.
In reaction 3, CN^- is both a strong base and a good nucleophile hence a mixture of substitution and elimination products are formed depending on the nature of the alkyl halide.