Given triangle $ABC$, the exterior angle at vertex $A$ is $134^\circ$ and the interior angle at vertex $B$ is $62^\circ$. We need to find the third interior angle of the triangle and the angle at which the angle bisectors of angles $A$ and $B$ intersect.

1. Problem Description

Given triangle $ABC$, the exterior angle at vertex $A$ is $134^\circ$ and the interior angle at vertex $B$ is $62^\circ$. We need to find the third interior angle of the triangle and the angle at which the angle bisectors of angles $A$ and $B$ intersect.

2. Solution Steps

First, we find the interior angle at vertex $A$. Let this be $\alpha$. Since the exterior angle at $A$ is $134^\circ$, we have

$180^\circ - \alpha = 134^\circ$

$\alpha = 180^\circ - 134^\circ = 46^\circ$

So, the interior angle at vertex $A$ is $46^\circ$.

Next, we find the interior angle at vertex $C$. Let this be $\gamma$. The sum of the interior angles of a triangle is $180^\circ$, so we have

$\alpha + \beta + \gamma = 180^\circ$

$46^\circ + 62^\circ + \gamma = 180^\circ$

$108^\circ + \gamma = 180^\circ$

$\gamma = 180^\circ - 108^\circ = 72^\circ$

So, the interior angle at vertex $C$ is $72^\circ$.

Now, we find the angle at which the angle bisectors of angles $A$ and $B$ intersect. Let $I$ be the point where the angle bisectors of angles $A$ and $B$ intersect. Let $\angle IAB$ be $\alpha/2$ and $\angle IBA$ be $\beta/2$. The angle bisector of $\angle A$ is $\frac{46^\circ}{2} = 23^\circ$ and the angle bisector of $\angle B$ is $\frac{62^\circ}{2} = 31^\circ$.

In triangle $AIB$, we have

$\angle IAB + \angle IBA + \angle AIB = 180^\circ$

$23^\circ + 31^\circ + \angle AIB = 180^\circ$

$54^\circ + \angle AIB = 180^\circ$

$\angle AIB = 180^\circ - 54^\circ = 126^\circ$

So, the angle at which the angle bisectors of angles $A$ and $B$ intersect is $126^\circ$.

3. Final Answer

The third angle of the triangle is $72^\circ$ and the angle at which the angle bisectors of angles $A$ and $B$ intersect is $126^\circ$.