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(A) The reaction proceeds via the formation of a carbocation intermediate. $1$. Protonation of the $-OH$ group by $HBr$ followed by the loss of $H_2O$

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$CH_3-C(CH_3)(Br)-CH_2-CH_3$

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(A) The reaction proceeds via the formation of a carbocation intermediate. $1$. Protonation of the $-OH$ group by $HBr$ followed by the loss of $H_2O$ leads to the formation of a secondary $(2^{\circ})$ carbocation: $CH_3-CH(CH_3)-CH^+-CH_3$. $2$. This $2^{\circ}$ carbocation undergoes a $1,2-H^-$ shift to form a more stable tertiary $(3^{\circ})$ carbocation: $CH_3-C^+(CH_3)-CH_2-CH_3$. $3$. Finally,the bromide ion $(Br^-)$ attacks the $3^{\circ}$ carbocation to form the major product: $CH_3-C(Br)(CH_3)-CH_2-CH_3$.

The major product of the following reaction is $:-$
$CH_3-CH(CH_3)-CH(OH)-CH_3 \xrightarrow{HBr} \text{Major Product}$

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The major product of the following reaction is $:-$ $CH_3-CH(CH_3)-CH(OH)-CH_3 \xrightarrow{HBr} \text{Major Product}$