Your friend was "half-way" right:
$(1)$ First finishing on your work, for $\,\tan\theta = \left(-\dfrac{3}{5}\right),\;$ if $\,\sin\theta < 0,\,$ we'd also need
$$\cos\theta = \dfrac{5}{\sqrt{34}} = \dfrac{5 \sqrt{34}}{34} > 0.\;$$ This makes sense since
$$\;\tan\theta = \dfrac{\sin\theta}{\cos\theta} = \dfrac{\frac{-3\sqrt{34}}{34}}{\frac{5\sqrt{34}}{34}} = \left(-\dfrac 35\right). $$
This, together with your work, gives that $P$ would then be in the fourth quadrant.
$(2)$ But... $P$ could also be located in Quadrant II: Since $\;\tan\theta = \dfrac{\sin\theta}{\cos\theta} = \left(-\dfrac 35\right)\lt 0,\;$ then exactly one of $\cos\theta, \sin \theta\,$ must be negative. $(1)$ gives one possible way this can happen.
But we might also have that $\cos \theta \lt 0, \sin\theta > 0$, putting $P$ in Quadrant II.
$$\;\tan\theta = \dfrac{\sin\theta}{\cos\theta} = \dfrac{\frac{3\sqrt{34}}{34}}{\frac{-5\sqrt{34}}{34}} = \left(-\dfrac 35\right).$$
Recall that $\,\sin\theta\,$ corresponds to the $\,y$-coordinate of $\,P\,$ in a unit circle, and $\,\cos \theta\,$ with its $\,x$-coordinate. When $\,x > 0\,$ and $\,y< 0,\,$ $\,(x, y)\,$ is in the Fourth Quadrant; when $\,x \lt 0,\;\text{and}\; y\gt 0,\;$ $(x, y)\,$ is in the Second Quadrant.
Hints:
What quadrant is the point $\;P = ({\bf x, y}) = (-3, 5)$ located?
Draw the right triangle that point $P$ makes with the $x$ axis - the length of the hypotenuse of the right triangle will equal $\;{\bf h} = \sqrt{(-3)^2 + 5^2} = \sqrt{34}$
Use SOH CAH TOA to unpack the definitions of $\tan \theta, \;\sin\theta,\;\cos\theta$:
$$\tan\theta = \frac{\text{opposite}}{\text{adjacent}} = \frac yx = \quad?\;$$
$$\sin\theta = \frac{\text{opposite}}{\text{hypotenuse}} = \frac yh = \quad?\;$$
$$\cos \theta= \frac{\text{adjacent}}{\text{hypotenuse}} = \frac{x}{h} = \quad?$$
Best Answer
For part a, $\sin\theta=\frac{3}{\sqrt{45}}=\frac{3}{3\sqrt{5}}=\frac{1}{\sqrt{5}}$ and since we know that the sine of an angle is the opposite over the hypotenuse we can imagine a triangle whose hypotenuse is $\sqrt{5}$ and whose opposite side is $1$. The Pythagorean theorem tells us that the adjacent side is then $\sqrt{\sqrt{5}^2-1^2}=\sqrt{5-1}=2$.
Since we are given that the point must be in the second quadrant, this gives a possible point of $(-2,1)$. Note that any positive multiple of this will also work, e.g. $(-4,2)$.
For part b, we use the fact that the cosine function is negative in the second and third quadrants and the tangent is positive in the first and third quadrants. Therefore, $\theta$ must be in the third quadrant for both conditions to be satisfied.