Find the values of real numbers $a,\ b$ for which the function $f(x)=a|\cos x|+b|\sin x|$ has local minimum at $x=-\frac{\pi}{3}$ and satisfies $\int_{-\frac{\pi}{2}}^{\frac{\pi}{2}}\{f(x)\}^{2}dx=2$.
1. **Given Function and Conditions:**
   We are given the function \( f(x) = a|\cos x| + b|\sin x| \) and need to find \(a\) and \(b\) such that:
   - \( f(x) \) has a local minimum at \( x = -\frac{\pi}{3} \)
   - \( \int_{-\frac{\pi}{2}}^{\frac{\pi}{2}} \{f(x)\}^2 \, dx = 2 \)

2. **Symmetry and Integral Simplification:**
   Since \( f(x) \) is an even function, we can simplify the integral:
   \[
   \int_{-\frac{\pi}{2}}^{\frac{\pi}{2}} \{f(x)\}^2 \, dx = 2 \int_{0}^{\frac{\pi}{2}} \{f(x)\}^2 \, dx = 2
   \]
   Therefore,
   \[
   \int_{0}^{\frac{\pi}{2}} \{f(x)\}^2 \, dx = 1
   \]

3. **Function Behavior in \([0, \frac{\pi}{2}]\):**
   In the interval \([0, \frac{\pi}{2}]\), \( f(x) = a \cos x + b \sin x \). We need to evaluate:
   \[
   \int_{0}^{\frac{\pi}{2}} (a \cos x + b \sin x)^2 \, dx
   \]

4. **Expanding the Integral:**
   \[
   \int_{0}^{\frac{\pi}{2}} (a \cos x + b \sin x)^2 \, dx = \int_{0}^{\frac{\pi}{2}} (a^2 \cos^2 x + 2ab \cos x \sin x + b^2 \sin^2 x) \, dx
   \]
   Using the trigonometric identities \( \cos^2 x = \frac{1 + \cos 2x}{2} \) and \( \sin^2 x = \frac{1 - \cos 2x}{2} \), we get:
   \[
   \int_{0}^{\frac{\pi}{2}} \left( a^2 \frac{1 + \cos 2x}{2} + 2ab \cos x \sin x + b^2 \frac{1 - \cos 2x}{2} \right) \, dx
   \]
   Simplifying further:
   \[
   \int_{0}^{\frac{\pi}{2}} \left( \frac{a^2 + b^2}{2} + \frac{a^2 - b^2}{2} \cos 2x + 2ab \cos x \sin x \right) \, dx
   \]

5. **Evaluating the Integral:**
   \[
   \int_{0}^{\frac{\pi}{2}} \frac{a^2 + b^2}{2} \, dx + \int_{0}^{\frac{\pi}{2}} \frac{a^2 - b^2}{2} \cos 2x \, dx + \int_{0}^{\frac{\pi}{2}} 2ab \cos x \sin x \, dx
   \]
   The integrals of \(\cos 2x\) and \(\cos x \sin x\) over \([0, \frac{\pi}{2}]\) are zero:
   \[
   \int_{0}^{\frac{\pi}{2}} \cos 2x \, dx = 0 \quad \text{and} \quad \int_{0}^{\frac{\pi}{2}} \cos x \sin x \, dx = 0
   \]
   Thus, we have:
   \[
   \int_{0}^{\frac{\pi}{2}} \frac{a^2 + b^2}{2} \, dx = \frac{a^2 + b^2}{2} \cdot \frac{\pi}{2} = 1
   \]
   Solving for \(a^2 + b^2\):
   \[
   \frac{\pi}{4} (a^2 + b^2) = 1 \implies a^2 + b^2 = \frac{4}{\pi}
   \]

6. **Local Minimum Condition:**
   For \( f(x) \) to have a local minimum at \( x = -\frac{\pi}{3} \), we need:
   \[
   f'\left(-\frac{\pi}{3}\right) = 0
   \]
   Since \( f(x) = a \cos x + b \sin x \) in \([0, \frac{\pi}{2}]\), we have:
   \[
   f'\left(x\right) = -a \sin x + b \cos x
   \]
   Setting \( f'\left(-\frac{\pi}{3}\right) = 0 \):
   \[
   -a \sin\left(-\frac{\pi}{3}\right) + b \cos\left(-\frac{\pi}{3}\right) = 0
   \]
   Using \(\sin\left(-\frac{\pi}{3}\right) = -\frac{\sqrt{3}}{2}\) and \(\cos\left(-\frac{\pi}{3}\right) = \frac{1}{2}\):
   \[
   -a \left(-\frac{\sqrt{3}}{2}\right) + b \left(\frac{1}{2}\right) = 0 \implies \frac{\sqrt{3}}{2} a + \frac{1}{2} b = 0 \implies \sqrt{3} a + b = 0 \implies b = -\sqrt{3} a
   \]

7. **Solving for \(a\) and \(b\):**
   Substituting \( b = -\sqrt{3} a \) into \( a^2 + b^2 = \frac{4}{\pi} \):
   \[
   a^2 + (-\sqrt{3} a)^2 = \frac{4}{\pi} \implies a^2 + 3a^2 = \frac{4}{\pi} \implies 4a^2 = \frac{4}{\pi} \implies a^2 = \frac{1}{\pi} \implies a = \pm \frac{1}{\sqrt{\pi}}
   \]
   Since \( a + b\sqrt{3} < 0 \) implies \( a < 0 \), we have:
   \[
   a = -\frac{1}{\sqrt{\pi}}
   \]
   And:
   \[
   b = -\sqrt{3} a = -\sqrt{3} \left(-\frac{1}{\sqrt{\pi}}\right) = \frac{\sqrt{3}}{\sqrt{\pi}}
   \]

The final answer is \( \boxed{ a = -\frac{1}{\sqrt{\pi}}, b = \frac{\sqrt{3}}{\sqrt{\pi}} } \)