Solving $T(n)=4T(n/2)-1$ without using the master theorem
The idea is to convert this recurrence to the simpler recurrence
$$ S(n) = 4S(n/2). $$
How do we do this? Let $S(n) = T(n) + f(n)$. Since
$$
S(n) = T(n) + f(n) = 4T(n/2) - 1 + f(n) = 4S(n/2) - 4f(n/2) - 1 + f(n),
$$
we need that
$$
f(n) = 4f(n/2) + 1.
$$
Luckily, $f(n) = -1/3$ is a solution. So $S(n) = T(n) - 1/3$ satisfies
$$ S(n) = 4S(n/2), \qquad S(4) = 14/3. $$
You take it from here.