1. Check statement (a)

We mix:

• $400\,\text{mL}$ of $0.1\,M\,H_2SO_4$
• $400\,\text{mL}$ of $0.1\,M\,NaOH$

First calculate moles:

$n(H_2SO_4)=0.4\times 0.1=0.04\text{ mol}$ $n(NaOH)=0.4\times 0.1=0.04\text{ mol}$

Since $H_2SO_4$ is dibasic, total acidic equivalents:

$0.04\times 2=0.08\text{ mol }H^+$

$NaOH$ provides:

$0.04\text{ mol }OH^-$

After neutralization, excess acid equivalents:

$0.08-0.04=0.04\text{ mol }H^+$

Total volume after mixing:

$V=0.4+0.4=0.8\,L$

So,

$[H^+]=\frac{0.04}{0.8}=0.05\,M$

Hence,

$\text{pH}=-\log(0.05)\approx 1.3$

So (a) is correct.

2. Check statement (b)

The ionic product of water is:

$K_w=[H^+][OH^-]$

This is an equilibrium constant, and equilibrium constants depend on temperature.

Hence (b) is correct.

3. Check statement (c)

Given a monobasic acid with

$K_a=10^{-5}$

For a weak monobasic acid $HA$ of concentration $C$, if its pH is 5:

$[H^+]=10^{-5}$

Degree of dissociation $\alpha$ is

$\alpha=\frac{[H^+]}{C}$

Also,

$K_a=\frac{C\alpha^2}{1-\alpha}$

If pH $=5$, then $[H^+]=10^{-5}$. Let us test whether $\alpha=50\%=0.5$ is possible.

Using

$K_a=\frac{C(0.5)^2}{1-0.5}=\frac{0.25C}{0.5}=0.5C$

Since $K_a=10^{-5}$,

$0.5C=10^{-5} \Rightarrow C=2\times 10^{-5}$

Then

$[H^+]=C\alpha=(2\times 10^{-5})(0.5)=10^{-5}$

So pH indeed becomes 5 and degree of dissociation is 50%.

Hence (c) is correct.

4. Check statement (d)

Common-ion effect is explained using suppression of ionization due to addition of a common ion, which is a direct application of Le Chatelier's principle.

Therefore the statement

Le Chatelier's principle is not applicable to common-ion effect

is false.

So (d) is incorrect.

5. Correct option

Correct statements are:

$\boxed{(a), (b), (c)}$

So the correct option is:

$\boxed{\text{B}}$