3. Divide the angle of $315^{\circ}$ into multiple angles (parts) such that each subsequent angle (part) is twice as large as the previous one. The size of each angle is a natural number in degrees. Find all solutions!
3. Let's divide the angle into two parts: $x$ and $2x$. Then we have:

$x + 2x = 315^{\circ}$

$3x = 315^{\circ}$

$x = 105^{\circ}$

$2x = 210^{\circ}$

$315^{\circ} = 105^{\circ} + 210^{\circ}$

Now, let's divide the angle into three parts: $x, 2x$, and $4x$. Then we have:

$x + 2x + 4x = 315^{\circ}$

$7x = 315^{\circ}$

$x = 45^{\circ}$

$2x = 90^{\circ}, \quad 4x = 180^{\circ}$

$315^{\circ} = 45^{\circ} + 90^{\circ} + 180^{\circ}$

Next, let's divide the angle into four parts: $x, 2x, 4x$, and $8x$. Then we have:

$x + 2x + 4x + 8x = 315^{\circ}$

$15x = 315^{\circ}$

$x = 21^{\circ}$

$2x = 42^{\circ}, \quad 4x = 84^{\circ}, \quad 8x = 168^{\circ}$

$315^{\circ} = 21^{\circ} + 42^{\circ} + 84^{\circ} + 168^{\circ}$

Now, let's divide the angle into five parts: $x, 2x, 4x, 8x$, and $16x$. Then we have:

$x + 2x + 4x + 8x + 16x = 315^{\circ}$

$31x = 315^{\circ}$

The number 315 is not divisible by 31, so it is not possible to divide the angle into five parts with the given condition that each subsequent part is twice the previous one.

Now, let's divide the angle into six parts: $x, 2x, 4x, 8x, 16x$, and $32x$. Then we have:

\[
\begin{aligned}
& x + 2x + 4x + 8x + 16x + 32x = 315^{\circ} \\
& 63x = 315^{\circ} \\
& x = 5^{\circ} \\
& 2x = 10^{\circ}, \quad 4x = 20^{\circ}, \quad 8x = 40^{\circ}, \\
& 315^{\circ} = 5^{\circ} + 10^{\circ} + 20^{\circ} + 40^{\circ} + 80^{\circ} + 160^{\circ}
\end{aligned}
\]

If we divide the angle into 7 or 8 parts with the given conditions, we get the equations $127x = 315^{\circ}$ and $255x = 315^{\circ}$, which do not have solutions in the set of natural numbers.

If we divide the angle into 9 or more parts with the given conditions, we find that the smallest part $x$ is less than $1^{\circ}$, so it cannot be a natural number.

Therefore, the angle of $315^{\circ}$ can be divided into 2, 3, 4, and 6 parts in the described manner.