Let $$f(z)=\frac{\cos(\frac{z}{6})+z\cdot\sin z}{\sin^2z}$$

$1)$

$\\$

Let $$g(z)=\cos(\frac{z}{6})+z\sin z,g(z)\neq 0,\therefore order =0.$$

Let $$h(z)=\sin^2 z, h(0)=0\\h'(z)=2\sin z\cos z=\sin 2z, h'(0)=0\\

h''(z)=2\cos 2z, h''z(0)\neq0, \therefore order=2.\\

2-0=2, \therefore \text{ it is a pole of order } 2.$$

$2)$

$\\$

Let $$z=3\pi + 6k\pi\\g(3\pi + 6k\pi)=0 \\ g'(z)=-\frac{1}{6}\sin(\frac{z}{6}) +z\cos z\\

g'(3\pi + 6k\pi)\neq 0,\therefore order =1.$$

Let $$h(z)=\sin^2 z, h(3\pi + 6k\pi)=0\\h'(z)=\sin 2z, h'(3\pi + 6k\pi)=0\\h''(z)=2\cos 2z, h''(3\pi + 6k\pi)\neq0\therefore order=2.\\2-1=1, \text{ order 1 simple pole}.$$

$3)$

$\\$

$$z=k\pi (k\neq 0), z\neq 3\pi + 6k\pi$$

Let $$g(z)=\cos(\frac{z}{6})+z\sin z,g(k\pi\neq 0),order=1$$

Let $$h(z)=\sin^2 z, h(k\pi)=0\\h'(z)=\sin 2z,h'(k\pi)=0\\h''(z)=2\cos 2z,h''(k\pi)\neq0, \therefore order=2$$

$$2-1=1, \text{ it is a simple pole}.$$