In Fig. 2,

Join the point T with O.

By theorem, we know that the lengths of tangents from the external point is equal. 
So, TP = TQ
In ΔTPQ,

TP = TQ

As two sides are equal, so the ΔTPQ is an isosceles triangle.

Here, OT is the bisector of ∠PTQ,

So, OT ⊥ PQ

∴ PR = RQ (perpendicular from centre to the chord bisects the chord)

So,

PR = QR = ½ PQ

= ½ (8)

= 4 cm

In ∆ OPR

OP² = OR² + PR²

5² = OR² + 4²
⇒ 25 = OR² + 16
⇒  9 = OR²

⇒ OR = 3 cm

In ∆ OPT

OT² = OP² + PT²

(TR + 3)² = 5² + PT²  …(1)

In ∆ PRT

PT² = TR² + RP²

PT² = TR² + 4² …(2)

Put the value of PT² in (1),

⇒ (TR + 3)² = 5² + TR² + 4²

⇒ TR² + 9 + 6TR = 25 + TR² + 16

⇒ 6TR + 9 = 25 + 16

⇒ 6TR  = 25 + 16 - 9

⇒ 6TR  = 32

From (2),

PT² = (16/3)² + 16

So, TP = 20/3 cm