The constant $a$ only describes attractive forces while the constant $b$ describes repulsive forces. So by definition $a$ cannot be negative. Can anyone tell me if this is correct?
It is correct.
Neither parameter $a$ or $b$ is regarded as negative, because a negative parameter would be opposite to its original meaning within the van der Waals theory. In agreement with this meaning, as pointed out by MaxW, when the theory is compared to data for real gases, the parameters are consistently positive.
As you correctly point out, the $b$ parameter encodes the repulsive interaction when gas particles come into close contact. Excluded volume refers to the same thing but can be encoded in different mathematical language by different models. The parameters in the van der Waals equation can be related to the second virial coefficient $B_2$ as follows:
$$ B_2 = b - \frac{a}{R T} \tag{1}$$
as can the parameters describing a square-well, Lennard-Jones, or other similar potential. All of these models can generate equivalent virial equations of state to first order, provided appropriate values are chosen for the coefficients (in some cases both attractive and repulsive terms are specified in the potential function). Note that equation (1) includes a second term, $-a/RT$ , that reflects the attractive nature of the interaction implied by the parameter $a$. When $B_2>0$ repulsions dominate. When $B_2<0$ attractions dominate.
So the pressure of real gas can be more as well as less than that if a gas behaving ideally depending on whether intermolecular forces are repulsive or attractive in nature.
Absolutely. The vdW equation can be written as
$$P = \frac{RT}{(V_\mathrm{m} -b)}- \frac{a}{V_\mathrm{m}^2}$$
Therefore for constant temperature and volume, the repulsive parameter $b$ results (intuitively) in a larger pressure, whereas the attractive parameter $a$ results in a reduced pressure. Equation (2) can be expanded as follows:
$$\begin{align} P &= \frac{RT/V_\mathrm{m}}{(1 -b/V_\mathrm{m})}- \frac{a}{V_\mathrm{m}^2}\ \\ &\approx \frac{RT}{V_\mathrm{m}}\left(1+\frac{b}{V_\mathrm{m}}\right)- \frac{a}{V_\mathrm{m}^2} \\ &= \frac{RT}{V_\mathrm{m}} + \frac{RTb-a}{V_\mathrm{m}^2} \end{align} $$
which shows that (to first order) when $RTb=a$ the gas behaves as if ideal. Consistently, in that case you also have that the second virial coefficient $B_2=0$. Depending on the relative magnitudes of $a$ and $b$ either attractive or repulsive interactions will predominate.
The answer to the question is given as True. So I want to understand what is wrong about my approach.
Nothing is wrong with your approach except. It is more usual that the repulsive term associated with $b$ is dominant, so that the pressure in a real gas is typically greater than in an ideal gas at the same $T$ and $V_m$. It is not impossible for the opposite to occur, though (see this problem, for instance, but beware the different naming convention of the parameters used there).
