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Answer

To determine the values of \(a\) and \(b\) such that the function \(f(x)\) is continuous at \(x = 0\), we need to analyze the left-hand limit (LHL) and right-hand limit (RHL) of \(f(x)\) as \(x\) approaches 0.

### Step 1: Calculate the Left-Hand Limit (LHL)

For \(x < 0\), the function is given by:
\[ f(x) = \frac{\sin((a+1)x) + \sin(2x)}{2x} \]
We need to find:
\[ \lim_{x \to 0^-} f(x) \]
Using the limit properties, we can separate the terms:
\[ \lim_{x \to 0^-} f(x) = \lim_{x \to 0^-} \left( \frac{\sin((a+1)x)}{2x} + \frac{\sin(2x)}{2x} \right) \]
Using the fact that \(\lim_{x \to 0} \frac{\sin(kx)}{kx} = 1\), we can rewrite:
\[ \lim_{x \to 0^-} \frac{\sin((a+1)x)}{2x} = \frac{(a+1)}{2} \quad \text{and} \quad \lim_{x \to 0^-} \frac{\sin(2x)}{2x} = 1 \]
Thus,
\[ \lim_{x \to 0^-} f(x) = \frac{(a+1)}{2} + 1 = \frac{(a+1) + 2}{2} = \frac{a + 3}{2} \]

### Step 2: Calculate the Right-Hand Limit (RHL)

For \(x > 0\), the function is given by:
\[ f(x) = \frac{\sqrt{x - bx^3} - \sqrt{x}}{bx^{5/2}} \]
We need to find:
\[ \lim_{x \to 0^+} f(x) \]
Factoring out \(\sqrt{x}\) from the numerator:
\[ = \frac{\sqrt{x(1 - bx^2)} - \sqrt{x}}{bx^{5/2}} = \frac{\sqrt{x}(\sqrt{1 - bx^2} - 1)}{bx^{5/2}} = \frac{\sqrt{1 - bx^2} - 1}{b x^2} \]
To evaluate this limit, we rationalize the numerator:
\[ \lim_{x \to 0^+} \frac{(\sqrt{1 - bx^2} - 1)(\sqrt{1 - bx^2} + 1)}{b x^2 (\sqrt{1 - bx^2} + 1)} = \lim_{x \to 0^+} \frac{-bx^2}{b x^2 (\sqrt{1 - bx^2} + 1)} = \lim_{x \to 0^+} \frac{-1}{\sqrt{1 - bx^2} + 1} \]
As \(x \to 0\), \(\sqrt{1 - bx^2} \to 1\):
\[ \lim_{x \to 0^+} f(x) = \frac{-1}{1 + 1} = -\frac{1}{2} \]

### Step 3: Set the Limits Equal for Continuity

For \(f(x)\) to be continuous at \(x = 0\):
\[ \lim_{x \to 0^-} f(x) = f(0) = \lim_{x \to 0^+} f(x) \]
This gives us:
\[ \frac{a + 3}{2} = b = -\frac{1}{2} \]
From \(b = -\frac{1}{2}\):
\[ \frac{a + 3}{2} = -\frac{1}{2} \]
Multiplying both sides by 2:
\[ a + 3 = -1 \implies a = -4 \]

### Step 4: Calculate \(|a + b|\)

Now we have:
\[ a = -4, \quad b = -\frac{1}{2} \]
Thus,
\[ a + b = -4 - \frac{1}{2} = -\frac{8}{2} - \frac{1}{2} = -\frac{9}{2} \]
Taking the absolute value:
\[ |a + b| = \left| -\frac{9}{2} \right| = \frac{9}{2} \]

### Final Answer
\[ \boxed{\frac{9}{2}} \]