Let's solve the equations one at a time.

### For Equation (a):
We have the equation:

\[
2^x + 2^{x+1} = 48
\]

We can rewrite \(2^{x+1}\) as \(2 \cdot 2^x\):

\[
2^x + 2 \cdot 2^x = 48
\]

This simplifies to:

\[
3 \cdot 2^x = 48
\]

Now, divide both sides by 3:

\[
2^x = \frac{48}{3} = 16
\]

Now, we know that \(16 = 2^4\). Thus, we can equate:

\[
2^x = 2^4
\]

Since the bases are the same, we can set the exponents equal to each other:

\[
x = 4
\]

### For Equation (b):
Now, for the equation:

\[
3^x + 2 - 3^{x+1} = 54
\]

We can rewrite \(3^{x+1}\) as \(3 \cdot 3^x\):

\[
3^x + 2 - 3 \cdot 3^x = 54
\]

This simplifies to:

\[
3^x + 2 - 3^{x+1} = 3^x + 2 - 3 \cdot 3^x = 54
\]

Combine like terms:

\[
-2 \cdot 3^x + 2 = 54
\]

Now, move the constant term to the right side:

\[
-2 \cdot 3^x = 54 - 2
\]

This simplifies to:

\[
-2 \cdot 3^x = 52
\]

Dividing by -2:

\[
3^x = \frac{52}{-2}
\]

This is:

\[
3^x = -26
\]

Since \(3^x\) cannot be negative for real \(x\), there is no real solution for this equation.

### Summary of Solutions:
- For (a), \(x = 4\)
- For (b), there are no real solutions.