= Solution
Keep $a$ general initially, since the final part does not explicitly fix it. For $-h\leq y\leq h$, where $h=\pi/6$, the <cumulative distribution function> is
$$
F_X(y)=\int_{-h}^y(ax+\cos x)\,dx=\frac a2(y^2-h^2)+\sin y+\frac12.
$$
The <normalization constant> in the <probability density function> of $Y$ must be
$$
M=\int_{-h}^h F_X(y)\,dy=\frac a2\left(\frac{2h^3}{3}-2h^3\right)+h=\boxed{h-\frac{2ah^3}{3}=\frac\pi6-\frac{a\pi^3}{324}.}
$$
Equivalently, the <integrated cumulative distribution function on a bounded interval> identity gives $M=h-\mathbb E[X]$. If the assumption $a=0$ from part (c) is retained, the requested special case is $\boxed{M=\pi/6}$. For the value $a=1$ in part (b), it is $M=\pi/6-\pi^3/324$.
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