Yes, you can use the known exact trigonometric values for $30^\circ$, $45^\circ$, and $90^\circ$ (as well as their complements and periodic properties) to solve all of these problems. Here's how:

1. Problem 1: $\cos(330^\circ)$
   Use the fact that $330^\circ = 360^\circ - 30^\circ$. This places the angle in the 4th quadrant where cosine is positive, and we know:
   $\cos(330^\circ) = \cos(30^\circ) = \frac{\sqrt{3}}{2}.$

2. Problem 2: $\sec(-315^\circ)$
   First, simplify the angle using the periodic property $360^\circ$:
   $-315^\circ + 360^\circ = 45^\circ.$
   Since $\sec(\theta) = \frac{1}{\cos(\theta)}$, we calculate:
   $\sec(-315^\circ) = \sec(45^\circ) = \frac{1}{\cos(45^\circ)} = \frac{1}{\frac{\sqrt{2}}{2}} = \sqrt{2}.$

3. Problem 3: $\sin(450^\circ)$
   Simplify $450^\circ$ using periodicity:
   $450^\circ - 360^\circ = 90^\circ.$
   Therefore, $\sin(450^\circ) = \sin(90^\circ) = 1.$

4. Problem 4: $\cot(120^\circ)$
   Rewrite $120^\circ$ as $180^\circ - 60^\circ$. This places the angle in the 2nd quadrant where tangent (and cotangent) is negative.
   Using $\cot(\theta) = \frac{\cos(\theta)}{\sin(\theta)}$, we know:
   $\cot(120^\circ) = -\cot(60^\circ) = -\frac{\cos(60^\circ)}{\sin(60^\circ)} = -\frac{\frac{1}{2}}{\frac{\sqrt{3}}{2}} = -\frac{1}{\sqrt{3}} = -\frac{\sqrt{3}}{3}.$

Final Results:

1. $\cos(330^\circ) = \frac{\sqrt{3}}{2}$
2. $\sec(-315^\circ) = \sqrt{2}$
3. $\sin(450^\circ) = 1$
4. $\cot(120^\circ) = -\frac{\sqrt{3}}{3}$

Would you like further explanations or visual aids for any of these?

Here are 5 related questions for further learning:

1. How do trigonometric functions behave in each quadrant?
2. What is the relationship between secant and cosine?
3. How can you quickly determine cotangent using reference angles?
4. Why are trigonometric functions periodic, and what are their periods?
5. How do complementary and supplementary angles relate to trigonometric functions?

Tip: Always simplify angles using periodicity ($360^\circ$ or $2\pi$) to make calculations easier!