For the 2nd one, you could always just draw a triangle. Since cot B = 4/5, the side opposite the angle is length 5 and the side adjacent to the angle B is 4.
Solving for the hypotenuse using c^2 = a^2 + b^2 , you get the hypotenuse as sqrt(41).
csc of an angle is equal to the hypotenuse/opposite, so csc B would be -sqrt(41)/5. (See edit).
Although I haven\'t done that sorta stuff in a while... I\'m still trying to think where the resrtiction 180<=B<=270 comes into play. If I think of it I\'ll let you know

Edit: Thought of it already

. The restriction comes into play if it is positive or negative. Using the CAST rule (dunno if you\'ve heard this or not... anyway 0-90 degress all is positive (the A) 90-180 only Sine is positive (S) 180-270 only tan is positive (T) and for 270-360 only cos is positive (C).) Anyway, since the angle is between 180 and 270, thats why cot B is positive (since cot is 1/tan). Therefore, the answer should be negative since csc (or 1/sin) is negative in that quadrant.