The user wants me to solve the following expression: $E = ({({\log _2}9)^2})^{{1 \over {{{\log }_2}({{\log }_2}9)}}} \times {(\sqrt 7 )^{{1 \over {{{\log }_4}7}}}}$

We can solve this by simplifying each part of the product separately.

Step 1: Simplify the first part of the expression

Let the first part be $A$. $A = ({({\log _2}9)^2})^{{1 \over {{{\log }_2}({{\log }_2}9)}}}$

Using the property of exponents $(x^m)^n = x^{m \times n}$, we get: $A = ({\log _2}9)^{2 \times {1 \over {{{\log }_2}({{\log }_2}9)}}}$ $A = ({\log _2}9)^{{2 \over {{{\log }_2}({{\log }_2}9)}}}$

To simplify this, let's use the exponent property $a^b = c^{b \log_c a}$. Let's choose the base $c=2$. Let $x = \log_2 9$. Then the expression for A is $x^{2/\log_2 x}$. Applying the property with $a = x$, $b = \frac{2}{\log_2 x}$ and $c = 2$: $A = 2^{\left( \frac{2}{\log_2 x} \right) \times (\log_2 x)}$ $A = 2^2$ $A = 4$

Step 2: Simplify the second part of the expression

Let the second part be $B$. $B = {(\sqrt 7 )^{{1 \over {{{\log }_4}7}}}}$

First, we simplify the exponent using the change of base formula for logarithms, ${\frac{1}{\log_b a}} = \log_a b$. ${1 \over {{{\log }_4}7}} = \log_7 4$

Substitute this back into the expression for $B$: $B = (\sqrt 7 )^{\log_7 4}$

We can write $\sqrt 7$ as $7^{1/2}$. $B = (7^{1/2})^{\log_7 4}$

Using the property of exponents $(x^m)^n = x^{m \times n}$: $B = 7^{\frac{1}{2} \log_7 4}$

Now, use the logarithm property $k \log_b a = \log_b(a^k)$: $B = 7^{\log_7 (4^{1/2})}$ $B = 7^{\log_7 (\sqrt 4)}$ $B = 7^{\log_7 2}$

Finally, using the fundamental property of logarithms $a^{\log_a x} = x$: $B = 2$

Step 3: Calculate the final value

The original expression is the product of $A$ and $B$. $E = A \times B$ $E = 4 \times 2$ $E = 8$

The value of the given expression is 8.