Let ℕ denote the set of all positive integers. Consider the sets A...

The condition $g(f(x)) = 2^x$ for all $x \in A$ implies that if $f(x_1) = f(x_2)$, then $g(f(x_1)) = g(f(x_2))$, which gives $2^{x_1} = 2^{x_2} \Rightarrow x_1 = x_2$.

Thus, $f$ must be an injective (one-to-one) function. Conversely, if $f$ is injective, such a function $g$ can always be constructed. Therefore, $T$ is the set of all injective functions $f: A \rightarrow B$ such that $f(2) \neq 2$ and $f(4) \neq 4$.

The total number of injective functions from $A$ to $B$ is $^{7}P_{5} = 7 \times 6 \times 5 \times 4 \times 3 = 2520$.

Let $E_2$ be the set of injective functions where $f(2) = 2$. The number of such functions is $^{6}P_{4} = 6 \times 5 \times 4 \times 3 = 360$.

Let $E_4$ be the set of injective functions where $f(4) = 4$. The number of such functions is $^{6}P_{4} = 360$.

Let $E_2 \cap E_4$ be the set of injective functions where both $f(2) = 2$ and $f(4) = 4$. The number of such functions is $^{5}P_{3} = 5 \times 4 \times 3 = 60$.

Using the Principle of Inclusion-Exclusion, the number of injective functions where $f(2) = 2$ or $f(4) = 4$ is:

$|E_2 \cup E_4| = |E_2| + |E_4| - |E_2 \cap E_4| = 360 + 360 - 60 = 660$

The number of elements in the set $T$ is the number of injective functions where $f(2) \neq 2$ and $f(4) \neq 4$, which is:

$|T| = 2520 - 660 = 1860$

Answer: $1860$