Answer 1

PT = TQ (Tangents drawn from a common point to the circle are equal)
So, ΔTPQ is isosceles with PT = TQ and TO is thus the angle bisector of ∠PTQ. 
So, OT ⊥⊥ PQ and thus, OT bisects PQ. (Perpendicular drawn from the center of the circle to the chord bisects the chord)
Thus, PR = RQ = 4 cm
Also, in ΔOPR,

OR =`sqrt("OP"^2-"PR"^2)`

`=sqrt(5^2 - 4^2)`

= 3 cm

In ΔTPR,

TR² + PR² = TP²

⇒ TR² + 4² = TP²

⇒ TR² + 16 = TP²

Let TP = x and TR = y

⇒ Y² + 16 = X² .....(1)

In Δ OPT,

TP² + OP² = OT²

⇒ TP² + 5² =  (TR + 3)²

⇒ X² + 25 = (Y + 3)².................(2)

Using (1) and (2) we get

y =`16/3`

putting this value of y in (1) we get

`(16/3)^2 + 16 = "x"^2`

⇒ `256/9 + 16 = "x"^2`

⇒ `(256+144)/9 ="x"^2`

⇒ `400/9 = "x"^2`

⇒ `"x" = 20/3`

Thus, the length of TP =`20/3`