Let's go through each part of this problem step-by-step:

(a) Verify that $f(x)$ is a PDF

To confirm that $f(x) = 10e^{-10x}$ is a probability density function (PDF), two criteria must be met:

1. Non-negativity: $f(x) \geq 0$ for all $x$ in the domain.
2. Integration to 1: The integral over the entire range of $f(x)$ must equal 1: $\int_{0}^{\infty} f(x) \, dx = 1.$

Since $f(x) = 10e^{-10x}$, we see that $f(x) \geq 0$ for $x \in (0, \infty)$, which satisfies the non-negativity condition.

Now, let's check if the integral over the range of $x$ equals 1: $\int_{0}^{\infty} 10e^{-10x} \, dx.$ To solve this, we use the fact that $\int e^{-ax} \, dx = \frac{1}{a}$ for a positive constant $a$. Here, $a = 10$: $\int_{0}^{\infty} 10e^{-10x} \, dx = 10 \cdot \frac{1}{10} = 1.$ Thus, $f(x)$ is indeed a valid PDF.

(b) Find $P(0 < X < 10)$ and $P(X > 5)$

To find probabilities involving $X$, we integrate the PDF over the relevant range.

For $P(0 < X < 10)$:

$P(0 < X < 10) = \int_{0}^{10} 10e^{-10x} \, dx.$ Using the same integration technique as before: $P(0 < X < 10) = \left[ -e^{-10x} \right]_{0}^{10} = -e^{-100} + e^{0} = 1 - e^{-100}.$ Since $e^{-100}$ is extremely small, $P(0 < X < 10) \approx 1$.

For $P(X > 5)$:

$P(X > 5) = \int_{5}^{\infty} 10e^{-10x} \, dx = \left[ -e^{-10x} \right]_{5}^{\infty} = 0 + e^{-50} = e^{-50}.$ So, $P(X > 5) = e^{-50}$, which is also very small.

(c) Find $\mu = E[X]$

The expected value $\mu = E[X]$ is given by $E[X] = \int_{0}^{\infty} x f(x) \, dx = \int_{0}^{\infty} x \cdot 10e^{-10x} \, dx.$ To solve this integral, we use integration by parts. Let $u = x$ and $dv = 10e^{-10x} \, dx$. Then $du = dx$ and $v = -e^{-10x}$, so [ E[X] = \left[ -x e^{-10x} \right]{0}^{\infty} + \int{0}^{\infty} e^{-10x} , dx. ] Evaluating the first term at $0$ and $\infty$, and solving the remaining integral, we find $E[X] = \frac{1}{10} = 0.1.$

(d) Find $E[(X - \mu)^2]$

To find the variance, $\sigma^2 = E[(X - \mu)^2]$, we use the formula [ \text{Var}(X) = E[X^2] - (E[X])^2. ] First, calculate $E[X^2]$: $E[X^2] = \int_{0}^{\infty} x^2 \cdot 10e^{-10x} \, dx.$ Using integration techniques, we can solve this integral to obtain