Answer
$3\frac{25}{128} \ in,3\frac{23}{128} \ in,\frac{1}{64}\ in$
Work Step by Step
The upper limit is the measurement + tolerance, that is,
$$
3\frac{3}{16}+\frac{1}{128}=3\frac{25}{128} \ in
.$$
The lower limit is the measurement - tolerance, that is,
$$
3\frac{3}{16}-\frac{1}{128}=3\frac{23}{128} \ in
.$$
The tolerance interval = upper limit-lower limit, that is,
$$
3\frac{25}{128} \ in-3\frac{23}{128} \ in=\frac{2}{128}\ in=\frac{1}{64}\ in
.$$
