1. To prove that $BA_1 \perp AC_1$, we note the following:

Since $A_1D \perp$ plane $ABC$ and $A_1D$ is contained in plane $ACC_1A_1$, plane $ACC_1A_1$ is perpendicular to plane $ABC$, and the intersection of plane $ACC_1A_1$ with plane $ABC$ is the line $AC$.

Because line $BC$ is contained in plane $ABC$ and $BC \perp AC$, it follows that $BC \perp$ plane $ACC_1A_1$.

Consequently, $BC \perp AC_1$.

Since $AA_1 = AC$, quadrilateral $ACC_1A_1$ is a rhombus, which means $AC_1 \perp A_1C$.

As $A_1C$ and $BC$ are contained in plane $A_1BC$ and they intersect at point $C$, we deduce that $AC_1$ is perpendicular to plane $A_1BC$.

Since $BA_1$ is contained in plane $A_1BC$, we conclude that $BA_1 \perp AC_1$.

2. To calculate the volume of tetrahedron $B_1-A_1DB$, we will use the fact that the volume of a tetrahedron is one sixth of the volume of its circumscribing prism.

The volume $V_{B_1-A_1DB}$ is equal to:

$$V_{B_1-A_1DB} = \frac{1}{2} V_{B-A_1B_1D} = \frac{1}{2} V_{D-A_1BB_1} = \frac{1}{2} \times \frac{1}{3} V_{C_1-A_1B_1B} = \frac{1}{2} \times \frac{1}{3} V_{B-A_1B_1C} = \frac{1}{6} V_{ABC-A_1B_1C_1}$$

Since the base area of the triangular prism $ABC-A_1B_1C_1$ is a right triangle with legs of length $2$, its area is $2 \times 2 \times \frac{1}{2} = 2$. The height of the prism is also $2$. Therefore, the volume of the prism is $2 \times 2 = 4$.

The volume of the tetrahedron $B_1-A_1DB$ is then:
$$\frac{1}{6} \times 4 = \frac{2}{3}$$

$\boxed{\frac{2}{3}}$
$$$$