Hand written answer is now replaced by typed answer, with an error in calculating#(dr)/(d(theta)# rectified:

First of all #5theta+(5pi)/3# is equivalent to #5theta + 2pi -pi/3# which is equivalent to #5theta -pi/3#. Hence #r=2theta -cos(5theta-pi/3)->(dr)/(d (theta))= 2+5sin(5theta-pi/3)#

Formula for the slope of a polar curve is #dy/dx=( (dr)/(d(theta)) sin theta + r cos theta)/((dr)/(d(theta)) cos theta -rsin theta)#

Slope is required at #theta= (-7pi)/4# which is equivalent to #pi/4 -2pi -> pi/4#. For this #theta#, sine and cosine values would be both #1/sqrt2#. Hence #dy/dx = ((dr)/(d(theta)) + r)/((dr)/(d(theta)) -r)#

Now at #theta=pi/4#, #r= pi/2 -cos((5pi)/4-pi/3)= pi/2 -cos((11pi)/12)# and #(dr)/(d(theta)) =2+5sin((11pi)/12)#

Substituting the values of #(dr)/(d(theta))# and r, slope would be

#(2 +pi/2 +5sin ((11pi)/12)- cos((11pi)/12))/(2-pi/2 +5sin((11pi)/12) +cos ((11pi)/12))#