Recall that converting from Cartesian to polar coordinates involves the identities
As a function in polar coordinates,
depends on 
, so you can write 
.
Differentiating the identities with respect to gives
The slope of the tangent line to
is given by
Given
, you have 
. So the tangent line to has a slope of
When
, the tangent line has slope
This line is tangent to the point
which in Cartesian coordinates is equivalent to 
, so the equation of the tangent line is
In polar coordinates, this line has equation
The tangent line passes through the y-axis when
, so the y-intercept is 
.
The vector from this point to the point of tangency on is given by the difference of the vector from the origin to the y-intercept (which I'll denote 
) and the vector from the origin to the point of tangency (denoted by 
). In the attached graphic, this corresponds to the green arrow.
The angle between this vector and the vector pointing to the point of tangency is what you're looking for. This is given by
The second problem is just a matter of computing the second derivative of with respect to 
and plugging in 
.
As a function in polar coordinates,
Differentiating the identities with respect to
The slope of the tangent line to
Given
When
This line is tangent to the point
In polar coordinates, this line has equation
The tangent line passes through the y-axis when
The vector from this point to the point of tangency on
The angle between this vector and the vector pointing to the point of tangency is what you're looking for. This is given by
The second problem is just a matter of computing the second derivative of