Consider the following function 
\[g(x)=(\alpha+|x|)^{2}e^{(5-|x|)^{2}}\]
[b]i)[/b] Find all the values of $\alpha$ for which $g(x)$ is continuous for all $x\in\mathbb{R}$

[b]ii)[/b]Find all the values of $\alpha$ for which $g(x)$ is differentiable for all $x\in\mathbb{R}$.
Given the function:
\[ g(x) = (\alpha + |x|)^2 e^{(5 - |x|)^2} \]

We need to find:
i) All values of \(\alpha\) for which \(g(x)\) is continuous for all \(x \in \mathbb{R}\).
ii) All values of \(\alpha\) for which \(g(x)\) is differentiable for all \(x \in \mathbb{R}\).

First, let's rewrite \(g(x)\) in piecewise form:
\[ g(x) = \begin{cases} 
(\alpha + x)^2 e^{(5 - x)^2} & \text{if } x > 0 \\
(\alpha - x)^2 e^{(5 + x)^2} & \text{if } x < 0 \\
\alpha^2 e^{5^2} & \text{if } x = 0 
\end{cases} \]

### Part i) Continuity

To check the continuity of \(g(x)\) at \(x = 0\), we need to ensure that:
\[ \lim_{x \to 0^+} g(x) = g(0) \quad \text{and} \quad \lim_{x \to 0^-} g(x) = g(0) \]

1. **Right-hand limit**:
   \[ \lim_{x \to 0^+} g(x) = \lim_{x \to 0^+} (\alpha + x)^2 e^{(5 - x)^2} \]
   As \(x \to 0^+\):
   \[ (\alpha + x)^2 \to \alpha^2 \]
   \[ e^{(5 - x)^2} \to e^{5^2} \]
   Therefore:
   \[ \lim_{x \to 0^+} g(x) = \alpha^2 e^{5^2} \]

2. **Left-hand limit**:
   \[ \lim_{x \to 0^-} g(x) = \lim_{x \to 0^-} (\alpha - x)^2 e^{(5 + x)^2} \]
   As \(x \to 0^-\):
   \[ (\alpha - x)^2 \to \alpha^2 \]
   \[ e^{(5 + x)^2} \to e^{5^2} \]
   Therefore:
   \[ \lim_{x \to 0^-} g(x) = \alpha^2 e^{5^2} \]

3. **Value at \(x = 0\)**:
   \[ g(0) = \alpha^2 e^{5^2} \]

Since:
\[ \lim_{x \to 0^+} g(x) = \lim_{x \to 0^-} g(x) = g(0) \]
\(g(x)\) is continuous at \(x = 0\) for all \(\alpha \in \mathbb{R}\).

### Part ii) Differentiability

To check the differentiability of \(g(x)\) at \(x = 0\), we need to ensure that:
\[ g'(0^+) = g'(0^-) \]

1. **Right-hand derivative**:
   \[ g'(x) = \frac{d}{dx} \left[ (\alpha + x)^2 e^{(5 - x)^2} \right] \]
   Using the product rule:
   \[ g'(x) = 2(\alpha + x) e^{(5 - x)^2} + (\alpha + x)^2 \cdot \frac{d}{dx} e^{(5 - x)^2} \]
   \[ \frac{d}{dx} e^{(5 - x)^2} = e^{(5 - x)^2} \cdot (-2(5 - x)) = -2(5 - x) e^{(5 - x)^2} \]
   Therefore:
   \[ g'(x) = 2(\alpha + x) e^{(5 - x)^2} - 2(\alpha + x)^2 (5 - x) e^{(5 - x)^2} \]
   As \(x \to 0^+\):
   \[ g'(0^+) = 2\alpha e^{5^2} - 10\alpha^2 e^{5^2} \]

2. **Left-hand derivative**:
   \[ g'(x) = \frac{d}{dx} \left[ (\alpha - x)^2 e^{(5 + x)^2} \right] \]
   Using the product rule:
   \[ g'(x) = 2(\alpha - x) e^{(5 + x)^2} + (\alpha - x)^2 \cdot \frac{d}{dx} e^{(5 + x)^2} \]
   \[ \frac{d}{dx} e^{(5 + x)^2} = e^{(5 + x)^2} \cdot 2(5 + x) = 2(5 + x) e^{(5 + x)^2} \]
   Therefore:
   \[ g'(x) = 2(\alpha - x) e^{(5 + x)^2} + 2(\alpha - x)^2 (5 + x) e^{(5 + x)^2} \]
   As \(x \to 0^-\):
   \[ g'(0^-) = 2\alpha e^{5^2} + 10\alpha^2 e^{5^2} \]

For \(g'(0)\) to exist:
\[ g'(0^+) = g'(0^-) \]
\[ 2\alpha e^{5^2} - 10\alpha^2 e^{5^2} = 2\alpha e^{5^2} + 10\alpha^2 e^{5^2} \]
\[ -10\alpha^2 e^{5^2} = 10\alpha^2 e^{5^2} \]
\[ -10\alpha^2 = 10\alpha^2 \]
\[ -20\alpha^2 = 0 \]
\[ \alpha = 0 \]

Therefore, \(g(x)\) is differentiable for all \(x \in \mathbb{R}\) if and only if \(\alpha = 0\).

The final answer is \( \boxed{ \alpha \in \mathbb{R} } \) for continuity and \(\alpha = 0\) for differentiability.