#r(theta)=3+8sin theta#

by converting into parametric equations,

#Rightarrow{(x(theta)=r(theta)cos theta=(3+8sin theta)cos theta),(y(theta)=r(theta)sin theta=(3+8sin theta)sin theta):}#

By differentiating with respect to #theta#,

#x'(theta)=8cos theta cdotcos theta+(3+8sin theta)cdot(-sin theta)#

#=8(cos^2theta-sin^2theta)-3sin theta#

#=8cos2theta-3sin theta#

by evaluating at #theta=pi/6#,

#Rightarrow x'(pi/6)=8cos(pi/3)-3sin(pi/6)=4-3/2=5/2#

By differentiating with respect to #theta#,

#y'(theta)=8cos theta cdot sin theta+(3+8sin theta)cdotcos theta#

#=16sin theta cos theta+3cos theta#

#=8sin2theta+3cos theta#

by evaluating at #theta=pi/6#,

#Rightarrow y'(pi/6)=8sin(pi/3)+3cos(pi/6)=4sqrt{3}+{3sqrt{3}}/2={11sqrt{3}}/2#

So, the slope #m# can be found by

#m={dy}/{dx}|_{theta=pi/6}={{dy}/{d theta}|_{theta = pi /6}}/{{dx]/{d theta}|_{theta = pi /6}}={y'(pi/6)}/{x'(pi/6)}={{11sqrt{3}}/{2}}/{{5}/{2}}={11sqrt{3}}/5#

The graph along with its tangent line at #theta=pi/6# looks like:

I hope that this was helpful.