First of all let me denote the sides with small letters #a#, #b# and #c#.

Let me name the angle between side #a# and #b# by #/_ C#, angle between side #b# and #c# by #/_ A# and angle between side #c# and #a# by #/_ B#.

Note:- the sign #/_# is read as "angle".

We are given with #/_C# and #/_A#. We can calculate #/_B# by using the fact that the sum of any triangles' interior angels is #pi# radian.

#implies /_A+/_B+/_C=pi#

#implies pi/12+/_B+(7pi)/12=pi#

#implies/_B=pi-(pi/12+(7pi)/12)=pi-(8pi)/12=pi-(2pi)/3=pi/3#

#implies /_B=pi/3#

It is given that side #b=2#.

Using Law of Sines

#(Sin/_B)/b=(sin/_C)/c#

#implies (Sin((pi)/3))/2=sin((7pi)/12)/c#

#implies (0.86602)/2=(0.96592)/c#

#implies 0.43301=0.96592/c#

#implies c=0.96592/0.43301#

#implies c=2.2307#

Therefore, side #c=2.2307#

Area is also given by

#Area=1/2bcSin/_A#

#implies Area=1/2*2*2.2307Sin((pi)/12)=2.2307*0.2588=0.5773# square units

#implies Area=0.5773# square units