Answer
a) $135.6^\circ$
b) See the figure below.
Work Step by Step
a)
To find the ball's initial angle in Nancy’s reference frame, we need to find the ball's velocity vector in Nancy’s reference frame.
$$v_{bN}=v_{bM}+v_{MN}$$
whereas $b\rightarrow$ ball, $N\rightarrow$ Nancy, $M\rightarrow$ Mike.
$$v_{bN}=v_{bM}-v_{NM}\tag 1$$
We know that ball's velocity relative to Mike is given by:
$$v_{bM}=22\cos63^\circ \;\hat i+22\sin63^\circ \;\hat j $$
We also know that Nancy velocity relative to mike is given by:
$$v_{NM}=30\cos 0^\circ \;\hat i+0\;\hat j $$
Plug these two formulas into (1):
$$v_{bN}=22\cos63^\circ \;\hat i+22\sin63^\circ \;\hat j -\left[30\cos 0^\circ \;\hat i+0\;\hat j \right]=-20.01\;\hat i+19.6\;\hat j$$
$$\boxed{v_{bN}=(-20.01\;{\rm m/s})\hat i+(19.6\;{\rm m/s})\hat j}$$
Thus,
$$\theta_{i,bN}=\tan^{-1}\left(\dfrac{v_{bN,y}}{v_{bN,x}}\right)=\tan^{-1}\left(\dfrac{ 19.6}{ -20.01}\right)=\bf -44.4^\circ$$
Ad since the angle in the third quadrant;
$$\theta_{i,bN}=180^\circ-44.4^\circ=\color{red}{\bf135.6}^\circ$$
b)
The ball's trajectory relative to Nancy is given by
$$\vec r=x_{bN}\;\hat i+y_{bN}\;\hat j$$
Thus,
$$\vec r=\left[v_{ix,bN}t\right]\;\hat i+\left[v_{iy,bN}t-\frac{1}{2}gt^2\right]\;\hat j$$
$$\vec r=\left[-20.02t\right]\;\hat i+\left[19.6t-4.9t^2\right]\;\hat j$$
Now we can draw the ball's trajectory relative to Nancy, as you see below.