Answer
$5\frac{13}{16} \ in,5\frac{11}{16} \ in,\frac{1}{8} \ in$
Work Step by Step
The upper limit is the measurement + tolerance, that is,
$$
5\frac{3}{4}+\frac{1}{16}=5\frac{13}{16} \ in
.$$
The lower limit is the measurement - tolerance, that is,
$$
5\frac{3}{4}-\frac{1}{16}=5\frac{11}{16} \ in
.$$
The tolerance interval = upper limit - lower limit, that is,
$$
5\frac{13}{16} \ in-5\frac{11}{16} \ in=\frac{2}{16} \ in=\frac{1}{8} \ in
.$$
