I have to evaluate the following:

tan((sin^-1)(2/7))

and

sin^-1 (sin(17pi/6))

Please help.

sin^-1 (2/7) means,

find the angle Ø , so that sinØ = 2/7

sketch a right-angled triangle with base angle Ø,
opposite side 2 and hyptenuse 7
by Pythagoras, adj^2 + 2^2 = 7^2
adjacent = √45 = 3√5

then tanØ = 2/(3√5)
thus tan(sin^-1 (2/7) = 2/(3√5)

for the second, I want you to really think about what you are looking at.
If you don't see it, follow my procedure.