Answer
a) $1.92\;\rm m/s$
b) $1.9 \;\rm m/s$
Work Step by Step
First, we need to identify the system here which we chose to be the man and the ball.
This system is isolated since the ice is frictionless which means that the momentum is conserved.
Thus,
$$\Delta p_i=\Delta p_f$$
Initially, the two of them are moving as one unit.
So,
$$(m_{man}+m_{ball})v_i=m_{man}v_{f,man}+m_{ball} v_{f,ball}\tag1$$
Solving for $v_{f,man}$;
$$v_{f,man}=\dfrac{(m_{man}+m_{ball})v_i-m_{ball} v_{f,ball}}{m_{man}} $$
a) When the ball's final speed after releasing is 15 m/s relative to the ground;
$$v_{f,man}=\dfrac{(70+0.45)(2)-(0.45) (15)}{70} $$
Noting that the two final velocities are in the same direction, to the right.
$$v_{f,man}=\color{red}{\bf 1.92}\;\rm m/s $$
b) When the ball's final speed after releasing is 15 m/s relative to the man. This means that the speed of the ball relative to the ground is given by
$$v_{f,ball}=v_{f,man}+15$$
Plugging into (1);
$$(m_{man}+m_{ball})v_i=m_{man}v_{f,man}+m_{ball} (v_{f,man}+15)$$
$$(m_{man}+m_{ball})v_i=m_{man}v_{f,man}+m_{ball} v_{f,man}+15m_{ball} $$
$$(m_{man}+m_{ball})v_i-15m_{ball}=v_{f,man}(m_{man}+m_{ball} )$$
Thus,
$$v_{f,man}=\dfrac{(m_{man}+m_{ball})v_i-15m_{ball}}{m_{man}+m_{ball}} $$
Plugging the known;
$$v_{f,man}=\dfrac{(70+0.45)(2)-(15)(0.45)}{70+0.45} $$
$$v_{f,man}=\color{red}{\bf 1.904}\;\rm m/s $$
