I can do your 2nd problem but not the 1st, for you've 
made a mistake in copying the first problem. Here's why:

sinq = 12/13, 3p/2 < q < 2p. Find sin2q, cos2q, tan2q?

3p/2 < q < 2p

tells us that q is in quadrant IV.
But the sine is negative in quadrant IV, yet you have 
sinq = 12/13, which is a positive number.
That cannot be.

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Here's your 2nd problem:

cosq = -1/4,cscq > 0, find sin2q,cos2q, tan2q

This tells us that q is in quadrant II,
because that is the only quadrant in which the cosine is negative and
the cosecant is positive.

So we draw an angle in Quadrant II to represent angle q 


Next we draw a perpendicular to the x axis.  Since the cosine
is , we make the x value -1, which is the numerator of 
 and make the hypotenuse or the r equal to the the 
denominator of  which is 4: 



Next we calculate the value of y by the Pythagorean
theorem equation







So we label the y-value with that




Now we can find sin2q using the
identity:

sin2q = 2sinqcosq

because we now know that sinq =  = , and cosq =  = .

sin2q = 2sinqcosq = 

Next we find cos2q using any one of
these three identities:

cos2q = cos2q - sin2q
cos2q = 2cos2q - 1
cos2q = 1 - 2sin2q.

I'll choose the second one, although any one will give you the
same answer:

cos2q = 2cos2q - 1
cos2q = 2 - 1
cos2q = 2 - 1
cos2q =  - 1
cos2q =  - 1
cos2q = 

Finally we can find tan2q from
the identity

tan2q = sin2q/cos2q

tan2q = 

Edwin