A brief hint of the expected answers. 
Note that on the examination itself you should explain what you are
doing, and not simply produce formulae.

Notation: subscripts are replaced by suffices, and superscripts are
denoted by means of the ^ character.

1a) p1 + p2 + p3 = 1

1b) p1 = 1-2p
    independent -> probability for both sites in state 1 is p1^2 etc
    -> E = - 2 p^2 - (1-2p)^2 = - 1 + 4p - 6p^2 per nearest neighbor 
    pair. Per site is z/2 times as much.

1c) S = -k (p1 log p1 + p2 log p2 + p3 log p3) =
    -k (2p log p + (1-2p)log(1-2p))

1d) T, p figure as independent parameters in E en S.
    Can be related by minimizing free energy F = E - T S.
    Hence eq (1)

1e) For p=1/3 both RHS and LHS vanish.
    To be expected on basis of permutation symmetry between the three
    states, since the symmetric point is expected to be a stationary point.
    
1f) LHS monotonically from +inf at p=0 to -inf at p=1/2
    RHS linearly from 1 naar -1/2 
    Them there must be two more solutions that for T->0 approach
    p->0 and p->1/2 respectively
    Last one meets p=1/3 as d/dp of RL and LL of
    (1) are equal -> kT(-2/(1-2p)-1/p) = -3z (op p=1/3)
    -> kT = z/3

1g) for high T the solution p=1/3 is unique minimum.
    for some lower T a new  min and max are born at  
    smaller p. They must have higher F . (because path from
    max to min is always decreasing).
    for yet lower T (kT=z/3) the max meets the min at p=1/3.
    At that T the other min must be lower (with argument above).
    Transition from one min to another, with max in between must
    be discontinuous  ->  first order.
    (note it is also given that at low T the unique pj is highest
    thus it follows p<1/3)

2a) Distance to transition line is certainly relevant.
    Since transition left and right of tricritical point are different,
    the parameter along transition line must also be relevant.
    yh is expected to be positive at the boundary of ferromagnetic
    phase.
    -> All three eigenvalues > 1 (i.e. all three y's positive)

2b) Flowlines all out of tricritical point.
    Initially tangential to transition line. (because y1>y2)



2c) Singular part of free energy:
     fs = U1^(d/y1) F[ U2 U1^(-y2/y1),  h U1^(-yh/y1) ]

    differentiate twice w.r.t. U1 -> alphat = 2 - d/y1 

2d) equivalent form of fs:
     fs = U2^(d/y2) Q[ U1 U2^(-y2/y2),  h U2^(-yh/y2) ]
    differentiate once w.r.t. h  ->  
    M ~ U2^(d/y2-yh/y2)
    Since T and Delta analytic in U1 and U2 the singular dependence
    on Delta has same power.

2e) use (at h=0) fs = U1^(d/y1) F[ U2 U1^(-y2/y1) ]
    F may be singular as U1->0 for non-zero U2.
    Singularity determined by fixed point toward which the 
    transition line flows.

2f) use (at h=0) fs = U1^(d/y1) F[ U2 U1^(-y2/y1) ]
    differentiate twice w.r.t. U1 -> alphat = 2 - d/y1 
    -> C  = U1^(d/y1-2) R[ U2 U1^(-y2/y1) ]
    (R is combination of F and its derivative)
    We know that U2 non-zero and that C ~ U1^(-alphac) ->
    U1^(-alphac) =  U1^(d/y1-2) [ U2 U1^(-y2/y1) ]^x
    -> -alphac  =  (d/y1-2) - x y2/y1
    -> x = (alphac y1 + d - 2 y1)/y2
    and A(U2) ~ U2^x