Solution:

Given: A,B,C are the vertices of a triangle and B,C lies on the line 3 x - 5 y = 1. Also, angle A = 45 degree.

To find: The slope of tangent at 'B' to the circumcircle of triangle ∆ABC.

Let's solve the problem step by step.

Step 1: Find the coordinates of point B and C.

Given, B and C lie on the line 3 x - 5 y = 1.

Let's assume y = t.

Then, 3 x - 5 t = 1.

Solving the above two equations, we get:

x = (5t + 1)/3

So, the coordinates of B and C are ((5t + 1)/3, t) and ((5s + 1)/3, s) respectively.

Step 2: Find the coordinates of point A.

Given, angle A = 45 degree.

Let's assume A lies on the line y = mx.

Then, B and C lie on the line 3 x - 5 y = 1.

So, the coordinates of A are (p, mp) where p = (5m + 5)/(3m^2 + 5).

Step 3: Find the circumcenter of triangle ∆ABC.

The circumcenter of triangle ∆ABC is the point of intersection of perpendicular bisectors of the sides.

Let's find the equation of the perpendicular bisector of BC.

The midpoint of BC is ((5t + 5s + 2)/6, (t + s)/2).

The slope of BC is (s - t)/((5s + 1)/3 - (5t + 1)/3) = (3s - 3t)/(5s - 5t) = -(3/5).

So, the equation of the perpendicular bisector of BC is:

(y - (t + s)/2) = (-5/3)(x - (5t + 5s + 2)/6)

Simplifying the above equation, we get:

5 x + 3 y - 5t - 5s - 2 = 0

Similarly, we can find the equation of the perpendicular bisector of AC.

The midpoint of AC is ((5t + 5p + 2)/6, (t + mp)/2).

The slope of AC is (mp - t)/((5p + 1)/3 - (5t + 1)/3) = (3mp - 3t)/(5p - 5t) = -(3/m).

So, the equation of the perpendicular bisector of AC is:

(y - (t + mp)/2) = (m/3)(x - (5t + 5p + 2)/6)

Simplifying the above equation, we get:

mx - 3y + 3mt + mp - 5p - 5t - 2 = 0

Solving the above two equations, we get the circumcenter of triangle ∆ABC as ((5t + 5p + 5s + 3)/6, (mp + t + s)/3).

Step 4: