a) It's enough to provide an example of such 5 integers. One example is the set $1, 2, 3, 4, 5$, whose arithmetic mean is $3$, and the largest common divisor is $1$.

b) Suppose that such numbers $a_1, a_2, a_3, a_4$ and $a_5$ exist, let $d$ be their largest common divisor, then these 5 integers can be rewritten as $a_i = d b_i$, $i = 1, 5$ and the equality in the statement is rewritten as:
$$\frac{a_1 + \cdots + a_5}{5} = 2d \Leftrightarrow d(b_1 + \cdots + b_5) = 10d \Leftrightarrow b_1 + \cdots + b_5 = 10.$$
Clearly numbers $b_1, \dots, b_5$ are distinct, so their smallest possible sum is $1 + 2 + 3 + 4 + 5 = 15 > 10$, this contradiction completes the proof.