For these kinds of problems, 1st you have to work backwards to find the value of θ, and then you can evaluate θ at the given angles.

So, starting with number 1, I think there was a little bit of a formatting issue, so I'm going to assume that's meant to say sinθ = -4/5

To get θ from this, take the arctangent of both sides, and you'll end up with -53° now, you're not quite done yet because you have to check the other condition. Any time you use an inverse trig function, there can be multiple potential answers. In this case, θ must be in quadrant IV so we add 360° to get our answer of 307°.

From here, you can plug 307° in to each of the trig functions to get the answer.

For number 2, the idea is effectively the same. I would however convert cscθ=2 into sinθ=1/2 by taking the reciprocal of both sides. From here we know θ is 30°, which is already in quadrant I, so you can go straight ahead and plug these in.

For number 3, before we do any work lets do a quick sign analysis. cotθ is positive, sinθ is negative. This can only happen in quadrant III so our answer should lie somewhere in there. Then, convert cotθ=1/5 to tanθ=5 (once again by taking the reciprocal of both sides). Then you can take the arctangent to solve for θ, and you'll get 78.69. This number is not in quadrant III, but as you know, tangent has a period of 180°, or in other words, any time you add or subtract 180° to an angle its tangent stays the same because of how the function repeats. 180°+78.69°=258.69° which you can then plug into the necessary trig functions.

Hope this helped!