Answer
a) $n_1=1.41$
b) No
c) $n=1.88$
Work Step by Step
a) $n_1=\frac{\sin(90^o)}{\sin(45^o)}=\sqrt{2}=1.41$
b) $\theta=\arcsin(\frac{n_2}{n_1})=\arcsin(\frac{1.33}{1.58})=57.3^o$
Since this is greater than $45^o$, the binoculars will not work underwater
c) $n=\frac{1.33}{\sin(45^o)}=1.88$