Suppose that $a$ and $b$ are real numbers such that $\sin(a)+\sin(b)=1$ and $\cos(a)+\cos(b)=\frac{3}{2}$. If the value of $\cos(a-b)$ can be written as $\frac{m}{n}$ for relatively prime positive integers $m$ and $n$, determine $100m+n$.

[i]Proposed by Michael Ren[/i]
1. Given the equations:
   \[
   \sin(a) + \sin(b) = 1
   \]
   \[
   \cos(a) + \cos(b) = \frac{3}{2}
   \]

2. Square both equations:
   \[
   (\sin(a) + \sin(b))^2 = 1^2 \implies \sin^2(a) + \sin^2(b) + 2\sin(a)\sin(b) = 1
   \]
   \[
   (\cos(a) + \cos(b))^2 = \left(\frac{3}{2}\right)^2 \implies \cos^2(a) + \cos^2(b) + 2\cos(a)\cos(b) = \frac{9}{4}
   \]

3. Add the squared equations:
   \[
   \sin^2(a) + \sin^2(b) + 2\sin(a)\sin(b) + \cos^2(a) + \cos^2(b) + 2\cos(a)\cos(b) = 1 + \frac{9}{4}
   \]

4. Use the Pythagorean identity \(\sin^2(x) + \cos^2(x) = 1\):
   \[
   (\sin^2(a) + \cos^2(a)) + (\sin^2(b) + \cos^2(b)) + 2\sin(a)\sin(b) + 2\cos(a)\cos(b) = 1 + \frac{9}{4}
   \]
   \[
   1 + 1 + 2\sin(a)\sin(b) + 2\cos(a)\cos(b) = \frac{13}{4}
   \]
   \[
   2 + 2\sin(a)\sin(b) + 2\cos(a)\cos(b) = \frac{13}{4}
   \]

5. Simplify the equation:
   \[
   2 + 2(\sin(a)\sin(b) + \cos(a)\cos(b)) = \frac{13}{4}
   \]
   \[
   2 + 2\cos(a-b) = \frac{13}{4}
   \]

6. Isolate \(\cos(a-b)\):
   \[
   2\cos(a-b) = \frac{13}{4} - 2
   \]
   \[
   2\cos(a-b) = \frac{13}{4} - \frac{8}{4}
   \]
   \[
   2\cos(a-b) = \frac{5}{4}
   \]
   \[
   \cos(a-b) = \frac{5}{8}
   \]

7. The value of \(\cos(a-b)\) is \(\frac{5}{8}\). To find \(100m + n\) where \(m = 5\) and \(n = 8\):
   \[
   100m + n = 100 \cdot 5 + 8 = 500 + 8 = 508
   \]

The final answer is \(\boxed{508}\)