To solve for the value of $\left(\frac{\operatorname{cosec} 1^{\circ}}{\alpha}\right)^2$, we begin by considering the expression for $\alpha$:

$ \alpha = \frac{1}{\sin 60^{\circ} \cdot \sin 61^{\circ}} + \frac{1}{\sin 62^{\circ} \cdot \sin 63^{\circ}} + \cdots + \frac{1}{\sin 118^{\circ} \cdot \sin 119^{\circ}} $

Using the identity for the product of sines:

$ \sin(x-y) = \sin x \cdot \cos y - \cos x \cdot \sin y $

we can transform each term:

$ \sin 1^{\circ} \cdot \alpha = \frac{\sin(61^{\circ} - 60^{\circ})}{\sin 60^{\circ} \cdot \sin 61^{\circ}} + \frac{\sin(63^{\circ} - 62^{\circ})}{\sin 62^{\circ} \cdot \sin 63^{\circ}} + \dots + \frac{\sin(119^{\circ} - 118^{\circ})}{\sin 118^{\circ} \cdot \sin 119^{\circ}} $

This simplifies to:

$ \sin 1^{\circ} \cdot \alpha = \cot 60^{\circ} - \cot 61^{\circ} + \cot 62^{\circ} - \cot 63^{\circ} + \cdots + \cot 118^{\circ} - \cot 119^{\circ} $

This forms a telescoping series, simplifying to:

$ \sin 1^{\circ} \cdot \alpha = \cot 60^{\circ} $

Therefore:

$ \alpha = \frac{\cot 60^{\circ}}{\sin 1^{\circ}} = \frac{1/\sqrt{3}}{\sin 1^{\circ}} = \frac{\operatorname{cosec} 1^{\circ}}{\sqrt{3}} $

Finally, we calculate:

$ \left(\frac{\operatorname{cosec} 1^{\circ}}{\alpha}\right)^2 = \left(\frac{\operatorname{cosec} 1^{\circ}}{\frac{\operatorname{cosec} 1^{\circ}}{\sqrt{3}}}\right)^2 = 3 $