#### Answer

(a) The sphere has symmetry with respect to the point $P$.
(b) The sphere has symmetry with respect to this line.

#### Work Step by Step

We can write the general equation of a sphere:
$(x-a)^2+(y-b)^2+(z-c)^2 = r^2$
where $(a,b,c)$ is the center of the sphere and $r$ is the radius
The equation of the sphere is: $(x-1)^2+(y-2)^2+(z+5)^2 = 49$
The center of the sphere is $(1,2,-5)$
(a) The only point that a sphere has symmetry with respect to is the sphere's center. Since the point $P = (1,2,-5)$ is the sphere's center, the sphere has symmetry with respect to the point $P$.
(b) A sphere has symmetry with respect to any line that passes through the sphere's center.
$l: (x,y,z) = (-5,4,-13)+ n(3,-1,4)$
If the line passes through the sphere's center, we can find the required value for $n$:
$x-coordinate: -5+n(3) = 1$
$3n = 6$
$n = 2$
We can find the point the line passes through when $n=2$:
$l: (x,y,z) = (-5,4,-13)+ n(3,-1,4)$
$(x,y,z) = (-5,4,-13)+ (2)(3,-1,4)$
$(x,y,z) = (-5+6,4-2,-13+8)$
$(x,y,z) = (1,2,-5)$
Since the line passes through the sphere's center, the sphere has symmetry with respect to this line.