Answer
Please see the work below.
Work Step by Step
(a) We know that
$m_1c_{copper}(T_1-T_c)=m_2c_{water}(T_c-T_2)$
We plug in the known values to obtain:
$0.5\times 386\times (80-T_c)=1\times 4184\times (T_c-10)$
$Tc=13.1C^{\circ}$
(b) As $\Delta S_{copper}=0.5\times 386\times \ln(\frac{13.1+273}{80+273})=-40.5\frac{J}{K}$
$\Delta S_{water}=1\times 4184\times \ln(\frac{13.1+273}{10+273})=44.1\frac{J}{K}$
Now $\Delta S_{system}=\Delta S_{copper}+\Delta S_{water}=-40.5+44.1=3.7\frac{J}{K}$
