It can be shown that the fourier transform of v(t) is given by: v(u)=(-4i/(u pi)) (1-cos(u pi^2))

Mathematica gives that directly

In[3]:= Clear[v]
v[t_] := Piecewise[{{0, -\[Pi] < t < -\[Pi]/2}, {-4 , -\[Pi]/2 < t < 0}, {4 , 0 < t < \[Pi]/2 }, {0, \[Pi]/2 < t < \[Pi]}}]

In[49]:= FourierTransform[v[t], t, u, FourierParameters -> {0, -2 \[Pi]}] // ExpToTrig // TrigReduce
Out[49]= (4 I (-1 + Cos[\[Pi]^2 u]))/(\[Pi] u)

The task a) is utterly easy with Mathematica (Why don't you use it?):

In[50]:= Clear[v]
v[u_] := (-4 I/(u \[Pi])) (1 - Cos[u \[Pi]^2])

In[52]:= ReIm /@ Join[{Limit[v[u], u -> 0]}, Table[v[u], {u, 1/(2 \[Pi]), 7/(2 \[Pi]), 1/(2 \[Pi])}]]
Out[52]= {{0, 0}, {0, -8}, {0, -8}, {0, -(8/3)}, {0, 0}, {0, -(8/5)}, {0, -(8/3)}, {0, -(8/7)}}

For task b) you explain how the formula for the fourier transform of v[t] (which is not a periodic function (because it is only defined for $-\pi \leq t \leq \pi$), but an odd function) coincides with the formulae for the computation of the fourier series coefficients $a_m, b_m$. To do that you use the identiy Exp[i x]=Cos[x]+ I Sin[x] as well as the fact that v[t] is odd and has finite support ( $-\pi \leq t \leq \pi$).

What you see is: all $a_m$ for the cosine fourier series are zero (the function is odd), also all real parts in part a) are zero. The $b_m$ you computed agree with the imaginary parts in part a) and represent the sine fourier series. This is the evidence to guide you through part b).