Answer
(a) $3.7\times 10^{-7}C $
(b) no
(c) The tension will be zero.
Work Step by Step
(a) We know that
$ T=\frac{Kq^2}{r^2}$
We plug in the known values to obtain:
$0.21N=\frac{9\times 10^9 Nm^2/C^2. q^2}{(7.6\times 10^{-2})^2}$
This simplifies to:
$ q=3.7\times 10^{-7}C $
(b) We know that the force responsible for the tension in the string depends only on the magnitude and not the sign of the charges. Thus, the given information is not useful in determining the sign of the charges.
(c) We know that even if the charges are positive or negative, a transfer of $+1\mu C $ will result the charges attaining the opposite signs. Thus, there is attractive force between the charges and the tension will be zero.
