Calculate the resistance of the lamp.

The voltage ($V$) across the lamp is $6.0 \, \text{V}$ and the current ($I$) through it is $0.53 \, \text{A}$.
Using Ohm's Law, $R = \frac{V}{I}$:

$$R = \frac{6.0 \, \text{V}}{0.53 \, \text{A}}$$ $$R \approx 11.32075 \, \Omega$$

Rounding to three significant figures:

$$R = 11.3 \, \Omega$$

The resistance is $\boxed{\text{11.3 } \Omega}$.

The voltage ($V$) is $6.0 \, \text{V}$ and the current ($I$) is $0.53 \, \text{A}$.
Using the formula for power, $P = VI$:

$$P = (6.0 \, \text{V}) \times (0.53 \, \text{A})$$ $$P = 3.18 \, \text{W}$$

The electrical power transferred is $\boxed{\text{3.18 W}}$.