From the reference Tangents with Polar Coordinates we obtain the equation:

#dy/dx = ((dr)/(d theta)sin(theta)+rcos(theta))/((dr)/(d theta)cos(theta)-rsin(theta))" [1]"#

The slope of the tangent line, #m_t,# is equation [1] evaluated at #theta = pi#:

#m_t = r/((dr)/(d theta))" [2]"#

The slope of the normal line, #m_n# is the negative reciprocal of equation [2]:

#m_n = -((dr)/(d theta))/r" [3]"#

We are given:

#r= f(theta)=theta- sin((5theta)/2-pi/2)+tan((2theta)/3-pi/2) #

Evaluated at #theta = pi#:

#r= pi + sqrt(3)/3 = (3pi+sqrt3)/3#

Compute #(dr)/(d theta)#:

#(dr)/(d theta) = 1 - 5/2cos((5theta)/2-pi/2)+2/3sec^2((2theta)/3-pi/2)#

Evaluated at #theta = pi#:

#-11/18#

#m_n = 11/(18pi+6sqrt3)#

Compute the Cartesian point from the polar point #(r,theta)=((3pi+sqrt3)/3,pi)#:

#x = rcos(theta)#

#x = (3pi+sqrt3)/3cos(pi)#

#x = -(3pi+sqrt3)/3#

#y = rsin(theta)#

#y = (3pi+sqrt3)/3sin(pi)#

#y = 0#

Use the point slope form, #y = m(x - x_1)+y_1# for the equation of a line:

#y = 11/(18pi+6sqrt3)(x - (-(3pi+sqrt3)/3))+ 0#

Simplify to slope-intercept form:

#y = 11/(18pi+6sqrt3)x +1/18#