check the picture below on the left-side.
so, negative angles move "clockwise", namely how the hands on a clock move, so -7π, is really -2π-2π-2π-π, which means, we go around the circle clockwise 3 times, as you see there in the green lines, and then we go a little bit more, namely -π, and that lands us on the -7π, or -π or the positive counterpart of π angle, same terminal point for all three.
and since tan(π) = 0, then tan(-7π) = 0, and likewise tan(-π) =0.
check the picture below on the right-side.
as you see there, the -3π/4 angle, will land there, namely is the same terminal point as the 5π/4 positive counterpart.
[tex] \bf csc\left(-\frac{3\pi }{4} \right)\implies csc\left(\frac{5\pi }{4} \right)\implies \cfrac{1}{sin\left(\frac{5\pi }{4} \right)}\implies \cfrac{1}{-\frac{\sqrt{2}}{2}}\implies -\cfrac{2}{\sqrt{2}}
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\stackrel{\textit{rationalizing the denominator}}{-\cfrac{2}{\sqrt{2}}\cdot \cfrac{\sqrt{2}}{\sqrt{2}}\implies -\cfrac{2\sqrt{2}}{(\sqrt{2})^2}}\implies -\cfrac{2\sqrt{2}}{2}\implies -\sqrt{2} [/tex]
