simple questions-please help

Q1)
i thnk u missed to mention Electric field direction. since u have not mentioned

I'm assuming that direction of E is along the vertical axis ( that is equitorial axis to dipole )
hence angle between the dipole and Electric field E is initially 90° which is considered as zero energy position \tau =pEsin\theta

let dW is the small amount of workdone in turning the dipole through a small angle dθ.
→ dW = \tau d\theta
→ dW = pEsin\theta d\theta

here we r to rotate 90° clockwise which means in the final position of dipole axis makes 0° with electric lines of force hence we r rotating dipole from zero enrgy position
( 90° ) to 0° position

hence W = \int_{0}^{W}{dW}\Rightarrow W=\int_{\pi /2}^{0}{pEsin\theta d\theta }=pE\int_{\pi /2}^{0}{sin\theta d\theta }
→ W = - pE

Q2)
shuldn't it be like this

F=qE ( due to electric field [downwards] )

T + qE = 2qE [ qEl = mv²/2 →mv²/l= 2qE ]
T = qE ??

2.

@manmay bhaiya it is given that system is kept in horizontal table.

ΔK.E= qEl = mv²/2
=> mv²/l= 2qE

let T be the tension at final position
T - qE=2qE
T= 3qE

but T + qE i guess that shuld be
as qE is towards O and T also [7]