= Solution
Put $L=5\pi/2$. The curve tangent is $\mathbf r'=(-b\sin\theta,b\cos\theta,c)$. For the first <line integral>, the <scalar triple product> identity gives $(\mathbf a\times\mathbf r)\cdot\mathbf r'=\mathbf a\cdot(\mathbf r\times\mathbf r')$. Its <cross product> is
$$
\mathbf r\times\mathbf r'=\bigl(bc(\sin\theta-\theta\cos\theta),\ -bc(\theta\sin\theta+\cos\theta),\ b^2\bigr).
$$
The needed <definite integrals> are
$$
\int_0^L(\sin\theta-\theta\cos\theta)d\theta
=\big[-\theta\sin\theta-2\cos\theta\big]_0^L=2-L,
$$
and $\int_0^L(-\theta\sin\theta-\cos\theta)d\theta=[\theta\cos\theta-2\sin\theta]_0^L=-2$. Hence
$$
\boxed{I_1=bc\left(2-\frac{5\pi}{2}\right)a_x-2bc\,a_y+\frac{5\pi b^2}{2}a_z.}
$$
For the second <line integral>, substitute the curve into $\mathbf v_2\cdot\mathbf r'$. The cubic terms combine to $b^3\cos\theta$, leaving
$$
\begin{aligned}
\mathbf v_2\cdot\mathbf r'
&=b^3\cos\theta+bc^2\theta^2\cos\theta+2bc^2\theta\sin\theta\\
&=\frac d{d\theta}\left[b\sin\theta\,(b^2+c^2\theta^2)\right].
\end{aligned}
$$
The <fundamental theorem of calculus> therefore gives
$$
\boxed{I_2=b\left(b^2+\frac{25\pi^2c^2}{4}\right).}
$$
Both calculations retain the full path from $\theta=0$ to $5\pi/2$, rather than inadvertently closing the <helix> after one turn.
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