The given function is a piecewise function defined as: $f(x) = \left\{ \begin{matrix} { - x - {\pi \over 2},} & {x \le - {\pi \over 2}} \\ { - \cos x} & { - {\pi \over 2} < x \le 0} \\ {x - 1} & {0 < x \le 1} \\ {\ln x} & {x > 1} \end{matrix} \right.$ We need to check the continuity and differentiability of the function at the points mentioned in the options.

Option A: f(x) is continuous at x = $-\pi$/2.

To check for continuity at $x = -\pi/2$, we need to evaluate the left-hand limit (LHL), the right-hand limit (RHL), and the value of the function at that point, i.e., $f(-\pi/2)$.

1. Function value: For $x = -\pi/2$, the function is defined as $f(x) = -x - \pi/2$. $f(-\pi/2) = -(-\pi/2) - \pi/2 = \pi/2 - \pi/2 = 0$

2. Left-hand limit (LHL): As $x$ approaches $-\pi/2$ from the left ($x < -\pi/2$), we use $f(x) = -x - \pi/2$. $LHL = \lim_{x \to (-\pi/2)^-} (-x - \pi/2) = -(-\pi/2) - \pi/2 = 0$

3. Right-hand limit (RHL): As $x$ approaches $-\pi/2$ from the right ($x > -\pi/2$), we use $f(x) = -\cos x$. $RHL = \lim_{x \to (-\pi/2)^+} (-\cos x) = -\cos(-\pi/2) = -\cos(\pi/2) = 0$

Since $LHL = RHL = f(-\pi/2) = 0$, the function $f(x)$ is continuous at $x = -\pi/2$. Thus, Option A is correct.

Option B: f(x) is not differentiable at x = 0.

First, let's check for continuity at $x = 0$.

1. Continuity at x = 0:

   • $f(0) = -\cos(0) = -1$.
   • $LHL = \lim_{x \to 0^-} f(x) = \lim_{x \to 0^-} (-\cos x) = -\cos(0) = -1$.
   • $RHL = \lim_{x \to 0^+} f(x) = \lim_{x \to 0^+} (x - 1) = 0 - 1 = -1$. Since $LHL = RHL = f(0)$, the function is continuous at $x=0$. Now we check for differentiability.

2. Differentiability at x = 0: We compute the left-hand derivative (LHD) and right-hand derivative (RHD) at $x=0$.

   • LHD: For $x < 0$, $f(x) = -\cos x$. So, $f'(x) = \sin x$. $LHD = f'(0^-) = \lim_{x \to 0^-} (\sin x) = \sin(0) = 0$
   • RHD: For $x > 0$, $f(x) = x - 1$. So, $f'(x) = 1$. $RHD = f'(0^+) = \lim_{x \to 0^+} (1) = 1$ Since $LHD \neq RHD$ ($0 \neq 1$), the function $f(x)$ is not differentiable at $x = 0$. Thus, Option B is correct.

Option C: f(x) is differentiable at x = 1.

First, let's check for continuity at $x = 1$.

1. Continuity at x = 1:

   • $f(1) = 1 - 1 = 0$.
   • $LHL = \lim_{x \to 1^-} f(x) = \lim_{x \to 1^-} (x - 1) = 1 - 1 = 0$.
   • $RHL = \lim_{x \to 1^+} f(x) = \lim_{x \to 1^+} (\ln x) = \ln(1) = 0$. Since $LHL = RHL = f(1)$, the function is continuous at $x=1$. Now we check for differentiability.

2. Differentiability at x = 1:

   • LHD: For $x < 1$, $f(x) = x - 1$. So, $f'(x) = 1$. $LHD = f'(1^-) = \lim_{x \to 1^-} (1) = 1$
   • RHD: For $x > 1$, $f(x) = \ln x$. So, $f'(x) = 1/x$. $RHD = f'(1^+) = \lim_{x \to 1^+} (1/x) = 1/1 = 1$ Since $LHD = RHD = 1$, the function $f(x)$ is differentiable at $x = 1$. Thus, Option C is correct.

Option D: f(x) is differentiable at x = -3/2.

We need to determine in which interval $x = -3/2$ lies. $x = -3/2 = -1.5$. We compare this value with the boundary point $-\pi/2 \approx -1.5708$. Since $-1.5708 < -1.5$, we have $-\pi/2 < -3/2$. Also, $-3/2 < 0$. So, the point $x = -3/2$ lies in the interval $(-\pi/2, 0]$.

In this interval, the function is defined as $f(x) = -\cos x$. The function $-\cos x$ is differentiable for all real numbers. Therefore, it is differentiable at any point in the open interval $(-\pi/2, 0)$. Since $x = -3/2$ is an interior point of the domain where $f(x) = -\cos x$, the function is differentiable at this point. The derivative is $f'(x) = \frac{d}{dx}(-\cos x) = \sin x$. At $x=-3/2$, the derivative $f'(-3/2) = \sin(-3/2) = -\sin(3/2)$ exists and is finite. Thus, Option D is correct.

Conclusion: All four options A, B, C, and D are correct.