Step 1: Analyze the given information and properties of tangents to a circle.
PA and PB are tangents from an external point P to a circle with center O.
The angle between the tangents, $\angle APB$, is given as $80^\circ$.
We need to find $\angle POA$.
Key properties of tangents from an external point:
The lengths of tangents from an external point to a circle are equal (PA = PB).
The line segment from the center to the external point (OP) bisects the angle between the tangents ($\angle APB$).
The radius drawn to the point of tangency is perpendicular to the tangent (OA $\perp$ PA and OB $\perp$ PB). This means $\angle OAP = \angle OBP = 90^\circ$.
Step 2: Use the property that OP bisects $\angle APB$.
Since OP bisects $\angle APB$, we have:
$\angle OPA = \frac{1}{2} \angle APB$
$\angle OPA = \frac{1}{2} \times 80^\circ$
$\angle OPA = 40^\circ$
Step 3: Consider triangle $\triangle OAP$.
We know that $\triangle OAP$ is a right-angled triangle because $\angle OAP = 90^\circ$ (radius is perpendicular to the tangent at the point of contact).
The sum of angles in a triangle is $180^\circ$. So, in $\triangle OAP$:
$\angle OAP + \angle OPA + \angle POA = 180^\circ$
Substitute the known angle values:
$90^\circ + 40^\circ + \angle POA = 180^\circ$
$130^\circ + \angle POA = 180^\circ$
Step 4: Solve for $\angle POA$.
$\angle POA = 180^\circ - 130^\circ$
$\angle POA = 50^\circ$
Alternatively, consider quadrilateral OAPB:
The sum of angles in a quadrilateral is $360^\circ$.
$\angle OAP + \angle APB + \angle PBO + \angle BOA = 360^\circ$
$90^\circ + 80^\circ + 90^\circ + \angle BOA = 360^\circ$
$260^\circ + \angle BOA = 360^\circ$
$\angle BOA = 360^\circ - 260^\circ = 100^\circ$
Since $\triangle OAP \cong \triangle OBP$ (RHS congruence criterion, as OA=OB (radii), OP=OP (common), $\angle OAP=\angle OBP=90^\circ$), OP also bisects $\angle AOB$.
Therefore, $\angle POA = \frac{1}{2} \angle BOA = \frac{1}{2} \times 100^\circ = 50^\circ$.
Step 5: Compare with the given options.
The calculated angle $\angle POA$ is $50^\circ$, which matches option (1).
\[
\mathbf{(1)} \quad 50^\circ
\]