Firstly, I would like to show $\left( 2+\sqrt3 \right) ^{2002} + \left( 2-\sqrt3 \right) ^{2002}=A$ is an even integer.
$\left( 2+\sqrt3 \right) ^{2002} + \left( 2-\sqrt3 \right) ^{2002}=(2 ^{2002}+2002.2^{2001}\sqrt3+\frac{2002.2001}{2}2^{2000}3+\frac{2002.2001.2000}{6}2^{1999}3\sqrt3+....) +(2 ^{2002}-2002.2^{2001}\sqrt3+\frac{2002.2001}{2}2^{2000}3-\frac{2002.2001.2000}{6}2^{1999}3\sqrt3+....)=2.(2 ^{2002}+\frac{2002.2001}{2}2^{2000}3+....)$
as shown above $\sqrt3$ terms will be zero after binom expansion, so A is an even integer.
$0<\left( 2-\sqrt3 \right) <1$ ,
$\frac{1}2 \left( 2+\sqrt3 \right) ^{2002} = \frac{A}2-\frac{1}2\left( 2-\sqrt3 \right) ^{2002}$
$A/2$ is an integer because A is an even integer and $0<\frac{1}2\left( 2-\sqrt3 \right) ^{2002}<1$ ,Thus
$ \lfloor \frac{1}2 \left( 2+\sqrt3 \right) ^{2002} \rfloor=\frac{A}2-1$
$ A \equiv y \pmod7 $ $ \left( 2+\sqrt3 \right) ^{2002} + \left( 2-\sqrt3 \right) ^{2002} \equiv y \pmod7 $
$ \left( (2+\sqrt3 \right)^2) ^{1001} + \left( (2-\sqrt3 \right)^2) ^{1001} \equiv y \pmod7 $
$ \left( (7+2\sqrt3 \right)) ^{1001} + \left( (7-2\sqrt3 \right)) ^{1001} \equiv y \pmod7 $
$ \left( (2\sqrt3 \right)) ^{1001} + \left( (-2\sqrt3 \right)) ^{1001} \equiv y \pmod7 $
$ \left( (2\sqrt3 \right)) ^{1001} - \left( (2\sqrt3 \right)) ^{1001} \equiv y \pmod7 $
$y=0$
and
$ A \equiv 0 \pmod7 $
$ \lfloor \frac{1}2 \left( 2+\sqrt3 \right) ^{2002} \rfloor=\frac{A}2-1$
$ \large \lfloor \frac{1}2 \left( 2+\sqrt3 \right) ^{2002} \rfloor= \frac{A}2-1 \equiv -1 \pmod7 $