(a)$f(x)=x^{3}-6x^{2}+9x+15$
so $f'(x)=3x^{2}-12x+9=0 \Rightarrow x=1,3$
Now $f''(x)=6x-12$
$f''(1) < 0,$ so $x = 1$ is point of local maxima, $f''(3) > 0,$ so $x = 3$ is point of local minima.
Also the end points $0$ and $6$ are critical points. $0$ is point of local minima, because it is to the left of $x = 1$ (which is point of maxima). Similarly $x = 6$ is point of local maxima.
(b) Since $x \cos x$ is an odd function, by the properties of definite integration, answer is $0$.