a) The numbers $137$ and $139$ are prime ($137 < 139 < 13^2$ and they have no positive divisors less than $13$), so each of them have $2^1$ divisors.
Since $138 = 2^1 \cdot 3^1 \cdot 23^1$, it results that $138$ has $2 \cdot 2 \cdot 2 = 2^3$ positive divisors.

b) Extend the sequence $137$, $138$, $139$ with four new numbers to obtain the $7$-element sequence $133$, $134$, $135$, $136$, $137$, $138$, $139$ such that each number has a number of positive divisors that is a power of $2$. Indeed, the numbers $136 = 2^3 \cdot 17$, $135 = 3^3 \cdot 5$, $134 = 2 \cdot 67$, $133 = 7 \cdot 19$ have, respectively, $8$, $8$, $4$, $4$ positive divisors.
We will prove that there are no $8$ consecutive numbers with the requested property. To this end, remark that among $8$ consecutive numbers there is one that leaves the remainder $4$ when divided by $8$, that is a number $a$ of the form $a = 8k+4 = 2^2(2k+1)$. In the prime factors decomposition of $a$, the prime $2$ appears at the second power and $2k+1$ is not a multiple of $2$. In consequence, the number of positive divisors of $a$ is divisible by $2+1=3$, so it is not a power of $2$.
Consequently, the largest sequence of consecutive positive integers with the numbers of their positive divisors powers of $2$ has $7$ elements.