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#
