For the compositions, just substitute and simplify:
#F(G(x))=F(1/(x+3))=\frac{2}{1/(x+3)-1}#
#=\frac{2x+6}{1-(x+3)}=(2x+6)/(-x-2)=-(2x+6)/(x+2)#
and
#G(F(x))=G(2/(x-1))=1/(2/(x-1)+3)#
#=(x-1)/(2+(3x-3))=(x-1)/(3x-1)#
For the domains, don't focus on the simplified formulas. Instead, either focus on the unsimplified formulas or the nature of the composition itself.
A number #x# will be in the domain of #F(G(x))# if and only if #x# is in the domain of #G# and #G(x)# is in the domain of #F#. The only number that is not in the domain of #G# is #x=-3#, and this number is also not in the domain of #F(G(x))#. Furthermore, if #x# is such that #G(x)=1#, then #x# is not in the domain of #F(G(x))#. Since #G(x)=1/(x+3)=1# if and only if #x+3=1\Leftrightarrow x=-2#, it follows that #x=-2# is also not in the domain of #F(G(x))#. Looking at the unsimplified formula #F(G(x))=\frac{2}{1/(x+3)-1}# is another way to help you see this.
In summary, the domain of #F(G(x))# is #\{x\in RR | x!=-3 \mbox{ and } x!=-2\}#
Looking at the formula #G(F(x))=1/(2/(x-1)+3)# and/or using the same logic as above helps us see that the domain of #G(F(x))# is #\{x\in RR| x!=1\mbox{ and } x!=1/3\}#.
The ranges are definitely trickier. For #F(G(x))=-(2x+6)/(x+2)# (when #x!=-3#), the graph of the expression #-(2x+6)/(x+2)# can be seen to pass through every horizontal line except for its horizontal asymptote at #y=-2#, so this number is not in the range. Also, since #x!=-3# is not in the domain of #F(G(x))#, the value of the expression #-(2x+6)/(x+2)# when #x=-3# is also excluded from the range. When #x=-3#, we get #-(2x+6)/(x+2)=-0/-1=0#. Therefore, the range of #F(G(x))# is #\{y\in RR | y!=-2\mbox{ and } y!=0\}#.
For #G(F(x))=(x-1)/(3x-1)# (when #x!=1#), the graph of the expression #(x-1)/(3x-1)# can be seen to pass through every horizontal line except for its horizontal asymptote at #y=1/3#, so this number is not in the range. Also, since #x!=1# is not in the domain of #G(F(x))#, the value of the expression #(x-1)/(3x-1)# when #x=1# is excluded from the range. When #x=1#, we get #(x-1)/(3x-1)=0/(-1)=0#. Therefore, the range of #G(F(x))# is #\{y\in RR | y!=1/3\mbox{ and } y!=0\}#.
