1. **State the problem:** Simplify the expression $$\sqrt{\left(4! + e^{\ln(32)}\right)^{-2} - \sin^2(0) + \frac{\log_2(64)}{\cos^2(0)}} + 2$$. 2. **Recall formulas and rules:** - Factorial: $4! = 4 \times 3 \times 2 \times 1 = 24$. - Exponential and logarithm: $e^{\ln(x)} = x$. - Trigonometric values: $\sin(0) = 0$, $\cos(0) = 1$. - Logarithm base 2: $\log_2(64) = 6$ because $2^6 = 64$. 3. **Evaluate inside the square root step-by-step:** - Calculate $4! = 24$. - Calculate $e^{\ln(32)} = 32$. - Sum inside parentheses: $24 + 32 = 56$. - Raise to power $-2$: $56^{-2} = \frac{1}{56^2} = \frac{1}{3136}$. - Calculate $\sin^2(0) = 0^2 = 0$. - Calculate $\cos^2(0) = 1^2 = 1$. - Calculate $\frac{\log_2(64)}{\cos^2(0)} = \frac{6}{1} = 6$. 4. **Combine all terms inside the square root:** $$\frac{1}{3136} - 0 + 6 = 6 + \frac{1}{3136} = \frac{6 \times 3136}{3136} + \frac{1}{3136} = \frac{18816 + 1}{3136} = \frac{18817}{3136}$$ 5. **Take the square root:** $$\sqrt{\frac{18817}{3136}} = \frac{\sqrt{18817}}{56}$$ 6. **Add 2 to the result:** $$2 + \frac{\sqrt{18817}}{56}$$ **Final answer:** $$2 + \frac{\sqrt{18817}}{56}$$