To find the time T when particles P and Q collide, we set up equations for their motions.

For particle P, starting from rest with acceleration 0.5 m s^-2, the distance covered is given by:

\(s_{AP} = \frac{1}{2} a t^2 = \frac{1}{2} \times 0.5 \times T^2 = 0.25 T^2\)

For particle Q, moving with constant speed 0.75 m s^-1, the distance covered is:

\(s_{BQ} = 0.75 T\)

Since the total distance between A and B is 10 m, we have:

\(s_{AP} + s_{BQ} = 10\)

Substituting the expressions for \(s_{AP}\) and \(s_{BQ}\), we get:

\(0.25 T^2 + 0.75 T = 10\)

This simplifies to the quadratic equation:

\(0.25 T^2 + 0.75 T - 10 = 0\)

Multiplying through by 4 to clear the fraction:

\(T^2 + 3T - 40 = 0\)

Factoring gives:

\((T + 8)(T - 5) = 0\)

Thus, \(T = 5\) (ignoring the negative solution \(T = -8\)).

For the speed of P immediately before the collision, we use:

\(v = u + at\)

Since \(u = 0\) and \(a = 0.5\), we have:

\(v = 0 + 0.5 \times 5 = 2.5 \text{ m s}^{-1}\)