To find the sum of the measures of the interior angles at vertices \( I \) and \( J \) in triangle \( HIJ \), we can use the property that the sum of all the interior angles of any triangle is always \( 180^\circ \).
Given that there is an exterior angle of \( 38^\circ \) at vertex \( H \), we know that the exterior angle is equal to the sum of the two opposite interior angles. Thus:
\[ \text{Exterior angle at } H = \angle I + \angle J \]
Substituting the known value:
\[ 38^\circ = \angle I + \angle J \]
Now, we want to find the sum of the interior angles at vertices \( I \) and \( J \):
\[ \angle I + \angle J = 38^\circ \]
However, this is the exterior angle based on an interior angle at \( H \). Since the angles of triangle \( HIJ \) sum to \( 180^\circ \), we can find the measures of angles at \( I \) and \( J \). Let \( \angle H = x \):
So:
\[ \angle H + \angle I + \angle J = 180^\circ \]
And since \( x \) is the exterior angle, we have:
\[ x = 180^\circ - 38^\circ = 142^\circ \]
Next, we can calculate \( \angle I + \angle J \):
\[ \angle I + \angle J = 180^\circ - 142^\circ \] \[ = 38^\circ \]
Thus, the sum of the measures of the interior angles at vertex \( I \) and vertex \( J \) is:
\[ \boxed{142^\circ} \]