As f(x) is continuous at x=2, we have
lim_(x->2^-)f(x) = lim_(x->2^+)f(x)
=> a(2^4)+5(2) = b(2^2)-3(2)
=> 16a+10 = 4b-6
=> a = 1/4b - 1
As f(x) is differentiable at x=2, the limit f'(2) = lim_(x->2)(f(x)-f(2))/(x-2) must exist. We can tell what the one sided limits will evaluate to by calculating the derivatives of the components of the piecewise defined functions on either side of 2.
lim_(x->2^-)(f(x)-f(2))/(x-2) = lim_(x->2^+)(f(x)-f(2))/(x-2)
=> 4a(2^3)+5 = 2b(2)-3
=> 32a+5 = 4b-3
Substituting in a = 1/4b-1, we have
32(1/4b-1) + 5 = 4b-3
=> 8b - 27 = 4b - 3
=> 4b = 30
:. b = 15/2