The power supply is switched on for 300s and Calculate (i) the charge, Q, that flows in the coil. (ii) electrical resistance, R, of the coil.

The power supply is switched on for $t = 300s$.
From the student's work in part (i), the current $I = 2.5A$ is used. We will assume this is the current flowing in the coil.

The formula relating current (I), charge (Q), and time (t) is $I = \frac{Q}{t}$.
To find the charge, we rearrange the formula to $Q = I \times t$.

The charge, Q, that flows in the coil is $\boxed{\text{750 C}}$.

The problem does not explicitly state the components of the coil. However, the student's work shows $R = R_1 + R_2 = 2.5 \Omega + 2.5 \Omega$. We will assume that the coil consists of two resistors, each with a resistance of $2.5 \Omega$, connected in series.
For resistors in series, the total resistance is the sum of individual resistances.

The electrical resistance, R, of the coil is $\boxed{\text{5 \Omega}}$.