Let us first consider the condition about the circle.
If $P(X,Y)$ is inside the circle, then there are no tangent lines from $P$ to the circle.
If $P(X,Y)$ is on the circle, then there is only one tangent line at $P$.
If $P(X,Y)$ is outside the circle, then there are two tangent lines from $P$ to the circle.
So, our condition about the circle is equivalent to that $P(X,Y)$ is either on or inside the circle, i.e.
$$X^2+Y^2\le 36\tag1$$
Next, let us consider the condition about the hyperbola.
Let $(p,q)$ be a point on the hyperbola $\frac{x^2}{25}-\frac{y^2}{16}=1$ where
$$\frac{p^2}{25}-\frac{q^2}{16}=1\iff 16p^2-25q^2=16\times 25\tag2$$
Since $y'=\frac{16x}{25y}$, the equation of the tangent line at $(p,q)$ is given by
$$y-q=\frac{16p}{25q}(x-p)$$
If $P(X,Y)$ is on the tanget line, we have
$$Y-q=\frac{16p}{25q}(X-p),$$
i.e.
$$25qY=16pX-(16p^2-25q^2),$$
Using here $(1)$ gives
$$25qY=16pX-16\times 25=16(pX-25)\tag3$$
Squaring the both sides of $(3)$, we get
$$25Y^2\times 25q^2=16^2(pX-25)^2$$
Using $(2)$, we have
$$25Y^2\times (16p^2-16\times 25)=16^2(pX-25)^2,$$
i.e.
$$(25Y^2-16X^2)p^2+(2\times 16\times 25\times X)p-25^2Y^2-16\times 25^2=0\tag4$$
Seeing this as a quadratic equation on $p$, our condition is equivalent to that $(4)$ has two distinct real solutions and the product of two real solutions is negative, i.e.
$$(2\times 16\times 25\times X)^2-4(25Y^2-16X^2)(-25^2Y^2-16\times 25^2)\gt 0\tag5$$and
$$\frac{-25^2Y^2-16\times 25^2}{25Y^2-16X^2}\lt 0\tag6$$
$(5)$ is equivalent to
$$16^2X^2+(25Y^2-16X^2)(Y^2+16)\gt 0\tag7$$
$(6)$ is equivalent to
$$25Y^2-16X^2\gt 0\tag8$$
Here, we can see that if $(8)$ holds, then $(7)$ holds.
So, our condition about the hyperbola is
$$(8)\iff (5Y-4X)(5Y+4X)\gt 0\tag9$$
Therefore, all we need is to find the options satisfying both $(1)$ and $(9)$.
$(a)(1,6)$ does not satisfy $(1)$.
$(b)(1,2)$ satisfies both $(1)$ and $(9)$.
$(c)(7,1)$ does not satisfy $(1)$.
$(d)(1,0.5)$ does not satisfy $(9)$.
It follows that $\color{red}{(b)}$ is the only correct option.
