Write 45^ in the form (1)/(sqrt(a)).

We know that $\sin 45^\circ = \frac{\sqrt{2}}{2}$.
To express this in the form $\frac{1}{\sqrt{a}}$, we can rationalize the numerator:

$$\sin 45^\circ = \frac{\sqrt{2}}{2} = \frac{\sqrt{2}}{\sqrt{4}}$$

To get 1 in the numerator, we can divide both the numerator and the denominator by $\sqrt{2}$:

$$\frac{\sqrt{2}}{\sqrt{4}} = \frac{\sqrt{2} \div \sqrt{2}}{\sqrt{4} \div \sqrt{2}} = \frac{1}{\sqrt{\frac{4}{2}}} = \frac{1}{\sqrt{2}}$$

So, $\sin 45^\circ = \frac{1}{\sqrt{2}}$.
Comparing this to $\frac{1}{\sqrt{a}}$, we find that $a=2$.
Since $a=2$ is a positive integer, this form is correct.

We can write $\sqrt{8}$ as:

$$\sqrt{8} = \sqrt{4 \times 2} = \sqrt{4} \times \sqrt{2} = 2\sqrt{2}$$

Using $\sin 45^\circ = \frac{1}{\sqrt{2}}$ and $\sqrt{8} = 2\sqrt{2}$:

$$\frac{\sqrt{8}}{1+\sin 45^\circ} = \frac{2\sqrt{2}}{1+\frac{1}{\sqrt{2}}}$$

$$\frac{2\sqrt{2}}{\frac{\sqrt{2}+1}{\sqrt{2}}} = 2\sqrt{2} \times \frac{\sqrt{2}}{\sqrt{2}+1}$$

Multiply the numerators:

$$2\sqrt{2} \times \sqrt{2} = 2 \times (\sqrt{2})^2 = 2 \times 2 = 4$$

So the expression becomes:

$$\frac{4}{\sqrt{2}+1}$$

Multiply the numerator and denominator by the conjugate of the denominator, which is $\sqrt{2}-1$:

$$\frac{4}{\sqrt{2}+1} \times \frac{\sqrt{2}-1}{\sqrt{2}-1}$$ $$= \frac{4(\sqrt{2}-1)}{(\sqrt{2})^2 - 1^2}$$ $$= \frac{4\sqrt{2}-4}{2-1}$$ $$= \frac{4\sqrt{2}-4}{1}$$ $$= 4\sqrt{2}-4$$

The simplified expression in surd form is:

$$4\sqrt{2}-4$$