For the first three problems, certain values of x need to be excluded.
(Problem #1) 2sin(x)/cos(3x) = tan(3x) - tan(x)
Let's try to prove that 2sin(x) = sin(3x) - tan(x)cos(3x).
Using the identities sin(3x) ? 3sin(x) - 4sin³(x) and cos(3x) ? 4cos³(x) - 3cos(x),
sin(3x) - tan(x)cos(3x) = 3sin(x) - 4sin³(x) - tan(x)(4cos³(x) - 3cos(x))
= sin(x)(3 - 4sin²(x) - (4cos²(x) - 3))
= sin(x)(6 - 4(sin²(x) + cos²(x)))
= 2sin(x).
(The above is simpler than using the standard identity that gives tan(3x) in terms of tan(x).)
(Problem #2) (1 - tan²((?/4)-x))/(1 + tan²((?/4)-x)) = sin(2x)
Multiplying the numerator and denominator of the left-hand side by cos²((?/4)-x)) gives
(cos²((?/4)-x)) - sin²((?/4)-x)))/(cos²((?/4)-x)) + sin²((?/4)-x))),
which equals cos(2(?/4 - x))/1, i.e., cos(?/2 - 2x), i.e., sin(2x).
(Problem #3) (1 + sin(2x) - cos(2x))/(1 + sin(2x) + cos(2x)) = tan(x)
Hint: use the identities cos(2x) = 1 - 2sin²(x) = 2cos²(x) - 1 and sin(2x) = 2sin(x)cos(x) to express the left-hand side in terms of sin(x) and cos(x), then factorize the numerator and denominator.
(Problem #4) sin(2x) + (1/sin(2x)) = 5/2; for 0 ? x <2?
This is an equation you have to solve, not an identity. Start by treating it as a quadratic equation in sin(2x). The quadratic factorizes easily.