Can anyone help me with Trigonometry?

Ten points for the best answer.

Find tan Θ/2 when 270º<Θ<360º and cosΘ is defined as indicated.
cosΘ = sqrt(7)/4

Find cosΘ/2 when 180º<Θ<270º and sinΘ is defined as indicated.
sinΘ = --sqrt(5)/3

Find the exact value of tan 13Π/8 using half-angle identities.

tan(t/2) =>
sqrt(sec(t/2)^2 - 1) =>
sqrt(1/cos(t/2)^2 - 1) =>
sqrt((1 - cos(t/2)^2) / cos(t/2)^2) =>
sqrt((1 - (1/2) * (1 + cos(t))) / ((1/2) * (1 + cos(t)))) =>
sqrt((2 - 1 - cos(t)) / (1 + cos(t))) =>
sqrt((1 - cos(t)) / (1 + cos(t))) =>
sqrt((1 - sqrt(7)/4) / (1 + sqrt(7)/4)) =>
sqrt((4 - sqrt(7)) / (4 + sqrt(7))) =>
sqrt((4 - sqrt(7))^2 / (16 - 7)) =>
(4 - sqrt(7)) / 3


sin(t) = -sqrt(5)/3
cos(t) = sqrt(1 - sin(t)^2)
cos(t) = sqrt(1 - 5/9)
cos(t) = sqrt(4/9)
cos(t) = +/- 2/3

Since we're in Q3, cos(t) = -2/3

cos(t/2) =>
sqrt((1/2) * (1 + cos(t))) =>
sqrt((1/2) * (1 - 2/3)) =>
sqrt(1/6) =>
sqrt(6)/6



tan(t) =>
tan(2 * t/2) =>
2 * tan(t/2) / (1 - tan(t/2)^2)

tan(13pi/4) = 2 * tan(13pi/8) / (1 - tan(13pi/8)^2)

13pi/4 is coterminal to 5pi/4

1 = 2 * tan(13pi/8) / (1 - tan(13pi/8)^2)
1 - tan(13pi/8)^2 = 2 * tan(13pi/8)
1 = tan(13pi/8)^2 + 2 * tan(13pi/8)
1 + 1 = tan(13pi/8)^2 + 2 * tan(13pi/8) + 1
2 = (tan(13pi/8) + 1)^2
+/- sqrt(2) = tan(13pi/8) + 1
tan(13pi/8) = -1 +/- sqrt(2)

13pi/8 is in Q2, so tan(13pi/8) < 0

-1 - sqrt(2)