#### Answer

$ \delta=0.4$ (or any smaller positive number)

#### Work Step by Step

When $x\in(2.6,3.8)$, then
$2.6 < x< 3.8$
$-0.4 < x-3 < 0.8$
and, we see from the graph that for these x,
$f(x)\in(1.5,2.5)$, that is,
$1.5 < f(x) < 2.5$
$-0.5 < f(x)-2 < 0.5$
$|f(x)-2| < 0.5\qquad(*)$
understanding that if x is within $0.4$ units from the left of $3$, or
within $0.8$ units to the right of $3$, the condition (*) is satisfied.
We will choose a $\delta$ within the interval $[0.4,0.8]$ such that
$-0.4\leq\ \ -\delta < x-1 < \delta \ \ \leq 0.8$
Since $\delta$ will describe a vicinity of equal distance from $3$ from either side, we choose the stricter (closer to $3$) vicinity.
We take $\delta=0.4$ (or any smaller positive number).
It then follows
$-0.4 < x-3 < 0.4\ \Rightarrow\ |x-3| <0.4$
which ensures that $x\in(2.6,3.8)$, which ensures
$|f(x)-2| < 0.5 $