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[tex]1) \ \large \boldsymbol {} \displaystyle 2 \frac{33}{40} +4\frac{3}{8} =2+\frac{33}{40}+4+\frac{3}{8}^{/5} =6+\frac{33+15}{40} = \\\\\\ 6\frac{48}{40} =7\frac{8}{40} =\boxed{7\frac{1}{5} } \\\\\\\\ 2) \ 2\frac{3}{8} +1\frac{1}{3} =2+1+\frac{3}{8}^{/3}+\frac{1}{3} ^{/8} =3+\frac{9+8}{24} =\boxed{3\frac{17}{24}} \rm \ cups[/tex]
