jmc

Let $x=\cos 1^{\circ }+i\sin 1^{\circ }$. Then from the identity$$\sin 1=\frac{x-\frac{1}{x}}{2i}=\frac{x^2-1}{2ix},$$we deduce that (taking absolute values and noticing $|x|=1$)$$|2\sin 1|=|x^2-1|.$$But because $\csc$ is the reciprocal of $\sin$ and because $\sin z=\sin (180^{\circ }-z)$, if we let our product be $M$ then$$\frac{1}{M}=\sin 1^{\circ }\sin 3^{\circ }\sin 5^{\circ }\dots \sin 177^{\circ }\sin 179^{\circ }$$$$=\frac{1}{2^{90}}|x^2-1||x^6-1||x^{10}-1|\dots |x^{354}-1||x^{358}-1|$$because $\sin$ is positive in the first and second quadrants. Now, notice that $x^2,x^6,x^{10},\dots ,x^{358}$ are the roots of $z^{90}+1=0.$ Hence, we can write $(z-x^2)(z-x^6)\dots (z-x^{358})=z^{90}+1$, and so$$\frac{1}{M}=\frac{1}{2^{90}}|1-x^2||1-x^6|\dots |1-x^{358}|=\frac{1}{2^{90}}|1^{90}+1|=\frac{1}{2^{89}}.$$It is easy to see that $M=2^{89}$ and that our answer is $2+89=\boxed{91}$.