Given a load, how do you calculate whether it is capacitive or inductive?
A capacitance \$C\$ has impedance

$$Z_{C} = \frac{1}{j\omega C}$$

But note that

$$\frac{1}{j} = -j$$

so equivalently

$$Z_{C} = -j\frac{1}{\omega C}$$

A capacitor's impedance therefore has a negative imaginary part.

An inductor \$L\$ has impedance

$$Z_{L} = j\omega L$$

and therefore has a positive imaginary part.

So if the load has a positive imaginary part then it behaves more like an inductor, and if it has a negative imaginary part then it behaves more like a capacitor.

If both inductors and capacitors are present then simply find the equivalent impedance of the load network. If the imaginary part of the equivalent impedance is positive then the load is inductive, if it is negative then it is capacitive, and if it is zero then it is resistive. An inductive load network has an overall higher impedance as the frequency increases even if there are capacitor(s) in the network, and a capacitive load network has an overall lower impedance as the frequency increases even if there are inductor(s) in the network.

It can be important to distinguish between the two cases in order to understand the frequency response of a circuit. For example, if a voltage amplifier has an inductive load then the amplifier will perform better at higher frequencies where the load has a higher impedance (the ideal load for a voltage amplifier is infinite in order to maximize the voltage gain). However, if the voltage amplifier's load is capacitive then it will perform better at lower frequencies since the load will then have a higher impedance at lower frequencies.