To solve this problem, we'll use the concepts of latent heat and electricity efficiency.

**Given:**
- Mass of aluminum per session = 15 tons = 15,000 kg
- Latent heat of fusion for aluminum, \( L = 4.00 \times 10^5 \, \text{J/kg} \)
- Efficiency of the furnace = 90%

**a) Calculate the heat energy (\( Q \)) needed to melt the aluminum:**

The formula for the heat energy required to melt a substance is:

\[ Q = m \times L \]

where
\( m \) = mass of the substance,
\( L \) = latent heat.

Plugging in the values:

\[ Q = 15,000 \, \text{kg} \times 4.00 \times 10^5 \, \text{J/kg} = 6.00 \times 10^9 \, \text{J} \]

**b) Calculate the electric power needed:**

The electric energy (\( W \)) considering efficiency is given by:

\[ W = \frac{Q}{\text{Efficiency}} \]

Efficiency is given as 90%, or 0.9 in decimal form. Therefore:

\[ W = \frac{6.00 \times 10^9 \, \text{J}}{0.9} = 6.67 \times 10^9 \, \text{J} \]

To convert Joules to kilowatt-hours (kWh):

\[ 1 \, \text{kWh} = 3.6 \times 10^6 \, \text{J} \]

\[ W_{\text{kWh}} = \frac{6.67 \times 10^9 \, \text{J}}{3.6 \times 10^6 \, \text{J/kWh}} = 1852.8 \, \text{kWh} \]

Rounded to whole units, \( W_{\text{kWh}} \approx 1853 \, \text{kWh} \).

The solution's expected answer might have been rounded differently, but the calculation should show:

- \( Q = 6 \times 10^9 \, \text{J} \)
- \( W \approx 1853 \, \text{kWh} \)