Let $x = \frac { \sqrt { 3 } + 1 } { \sqrt { 3 } - 1 }$ and $y = \frac { \sqrt { 6 } - \sqrt { 2 } } { \sqrt { 6 } + \sqrt { 2 } }$.
(1) We have $x = \mathbf { A } + \sqrt { \mathbf { B } }$ and $y = \mathbf { C } - \sqrt { \mathbf { C } }$. Hence we have
$$x + y = \mathbf { E } , \quad x y = \mathbf { F } , \quad \frac { 1 } { x ^ { 2 } } + \frac { 1 } { y ^ { 2 } } = \mathbf { G H } .$$
Also we have
$$5 \left( x ^ { 2 } - 4 x \right) + 3 \left( y ^ { 2 } - 4 y + 1 \right) = \square \mathbf { I J } .$$
(2) The values of the integers $m$ and $n$ such that $\frac { m } { x } + \frac { n } { y } = 4 + 4 \sqrt { 3 }$ are
$$m = \mathbf { K L } , \quad n = \mathbf { M } .$$