Define fundamental quantities and give examples. 2. Determine the dimensions of 'a' and 'b' in Van der Waals' equation.

For quantities to be subtracted or added, they must have the same dimensions. In the term $(V - b)$, $V$ is volume.
Therefore, the dimension of $b$ must be the same as the dimension of $V$.
The dimension of volume $V$ is $[L^3]$.
Thus, the dimension of $b$ is $\boxed{[L^3]}$.

In the term $\left(P + \frac{a}{V^2}\right)$, $P$ is pressure. For quantities to be added, they must have the same dimensions.
Therefore, the dimension of $\frac{a}{V^2}$ must be the same as the dimension of $P$.
First, find the dimension of pressure $P$.
Pressure $P = \frac{\text{Force}}{\text{Area}}$.
Dimension of Force = $[M][L][T^{-2}]$ (Mass $\times$ Acceleration)
Dimension of Area = $[L^2]$
Dimension of $P = \frac{[M][L][T^{-2}]}{[L^2]} = [M][L^{-1}][T^{-2}]$.

Dimension of $\frac{a}{V^2}$ = Dimension of $P$
$\frac{\text{Dimension of } a}{(\text{Dimension of } V)^2} = [M][L^{-1}][T^{-2}]$
We know Dimension of $V = [L^3]$, so $(\text{Dimension of } V)^2 = ([L^3])^2 = [L^6]$.
$\frac{\text{Dimension of } a}{[L^6]} = [M][L^{-1}][T^{-2}]$

Dimension of $a = [M][L^{-1}][T^{-2}] \times [L^6]$
Dimension of $a = [M][L^{(-1+6)}][T^{-2}]$
Thus, the dimension of $a$ is $\boxed{[M][L^5][T^{-2}]}$.