Let $f$ be a function such that $f(0)=1$, $f^\prime (0)=2$, and \[f^{\prime\prime}(t)=4f^\prime(t)-3f(t)+1\] for all $t$.  Compute the $4$th derivative of $f$, evaluated at $0$.
1. We are given the function \( f \) with initial conditions \( f(0) = 1 \) and \( f'(0) = 2 \). Additionally, we have the second derivative given by:
   \[
   f''(t) = 4f'(t) - 3f(t) + 1
   \]
   We need to find the fourth derivative of \( f \) evaluated at \( t = 0 \).

2. First, we compute \( f''(0) \) using the given formula:
   \[
   f''(t) = 4f'(t) - 3f(t) + 1
   \]
   Substituting \( t = 0 \):
   \[
   f''(0) = 4f'(0) - 3f(0) + 1 = 4 \cdot 2 - 3 \cdot 1 + 1 = 8 - 3 + 1 = 6
   \]

3. Next, we need to find \( f'''(0) \). To do this, we differentiate the given second derivative formula:
   \[
   f''(t) = 4f'(t) - 3f(t) + 1
   \]
   Differentiating both sides with respect to \( t \):
   \[
   f'''(t) = 4f''(t) - 3f'(t)
   \]
   Substituting \( t = 0 \):
   \[
   f'''(0) = 4f''(0) - 3f'(0) = 4 \cdot 6 - 3 \cdot 2 = 24 - 6 = 18
   \]

4. Finally, we need to find \( f^{(4)}(0) \). Using the formula for the third derivative:
   \[
   f'''(t) = 4f''(t) - 3f'(t)
   \]
   We differentiate again:
   \[
   f^{(4)}(t) = 4f'''(t) - 3f''(t)
   \]
   Substituting \( t = 0 \):
   \[
   f^{(4)}(0) = 4f'''(0) - 3f''(0) = 4 \cdot 18 - 3 \cdot 6 = 72 - 18 = 54
   \]

The final answer is \( \boxed{54} \)