Explanation

(a) Since resistance connected in arms $CE,\ ED,\ CF$ and $FD$ will form a balanced Wheatstone bridge, therefore, the resistance of arm $EF$ becomes ineffective. Now resistance of arm $CED\ $or$\ CFD = 2 + 2 = 4\Omega$. Effective resistance of these two parallel arm$= \frac{4 \times 4}{4 + 4} = 2\Omega$
Now resistance of arm $ACDB = 2 + 2 + 2 = 6\Omega$, is in parallel with resistance arm $AB = 2\Omega$. Thus, effective resistance between $A$ and $B$
$= \frac{6 \times 2}{6 + 2} = \frac{3}{2}\Omega$