1,3,4 can be solved by Master's Theorem, not sure about 2
1.
$f(n)= 2^n , a=1, b=2$
$f(n)= \Omega\left(n^{\log_b a + \epsilon }\right)$. Any $\epsilon \in \mathbb{N}$ would do as exponential function grows faster than polynomial.Now, for Master theorem case 3 we need to see regularity condition also, which is
$$af\left(\frac{n}{b}\right) \leq c f(n), c < 1$$
$\implies 2^{\frac{n}{2}} \leq c 2^n$
$c = 0.5$ works here and so we can apply Master theorem Case 3
$$T(n) = \Theta(f(n)) = \Theta\left( 2^n \right)$$
3.
$f(n) = n! , a = 16, b = 4$
$f(n) = \Omega\left(n^{\log_b a + \epsilon }\right)$. Any $\epsilon$ would do as factorial grows at a faster rate than polynomial. So, we have Case 3 of Master theorem which also requires the regularity check.
$$af\left(\frac{n}{b}\right) \leq c f(n), c < 1$$
$\implies 16 . {\frac{n}{4}} ! \leq c n!$
is surely true for $c = \frac{1}{16}$. So, we can apply Case 3 of Master theorem and so, $$T(n) = \Theta(f(n)) = \Theta( n! )$$
4.
$f(n) = \log n , a = √2, b = 2$
$f(n) = O\left(n^{\log_b a - \epsilon}\right )$ take $\epsilon = 0.1$
Master theorem Case 1 and hence, $$T(n) = \Theta\left(n^{\log_2 \sqrt2} \right)= \Theta( \sqrt n )$$
For 2 we can't apply Master theorem- why because $n^{\log_b a} = n^{\log_2 2} = n$. Now $n/ \lg n = O(n)$ is true but we can't write $n/\lg n = O\left(n^{1 - \epsilon}\right)$ for any positive epsilon. Thus we can't apply master theorem. Solution can be obtained using Extended Master theorem or solving recurrence directly. You can see the below link.
https://gateoverflow.in/10619/recurrence