Answer
(a) $ T(r) = \frac{mv_1^2r_1^2}{r^3} $
(b) $ W_T = \frac{mv_1^2}{2}r_1^2 \Big( \frac{1}{r_2^2} - \frac{1}{r_1^2} \Big) $
(c) $ \Delta K = W_T $.
Work Step by Step
(a) Angular momentum of the block is conserved, since there is no net torque:
$$ L_1 = L \Rightarrow I \omega_1 = I \omega \Rightarrow v_1r_1 = vr \Rightarrow v= \frac{v_1r_1}{r} $$
For the block:
$$ \Sigma F = T = ma_{\mathrm{rad}} = \frac{mv^2}{r} = \frac{mv_1^2r_1^2}{r^3} = T(r)$$
(b) The displacement of the box is in the same direction as the tension force doing work (towards the hole), but our displacement magnitude $r_2 - r_1$ is negative, so if we want the work done by the tension to be positive (like it should) we need to add a minus-sign into the work equation:
$$ W_T = -\int_{r_1}^{r_2} T(r) \, \mathrm{d}r = -mv_1^2r_1^2 \int_{r_1}^{r_2} \frac{1}{r^3} \, \mathrm{d}r = -mv_1^2r_1^2 \bigg[ -\frac{1}{2r^2} \bigg]_{r_1}^{r_2} = $$
$$ -mv_1^2r_1^2 \bigg( -\frac{1}{2r_2^2} + \frac{1}{2r_1^2} \bigg) = \frac{mv_1^2}{2}r_1^2 \bigg( \frac{1}{r_2^2} - \frac{1}{r_1^2} \bigg) $$
(c) $$ \Delta K = \frac{mv_2^2}{2} - \frac{mv_1^2}{2} = \frac{mv_1^2r_1^2}{2r_2^2} - \frac{mv_1^2}{2} = \frac{mv_1^2}{2}r_1^2 \bigg( \frac{1}{r_2^2} - \frac{1}{r_1^2} \bigg) $$
