Minimum number of Poles and zero of transfer function H(z)?
Restricting the discussion to LTI systems with rational transfer functions, there are various conditions on the placement of poles and zeros  of $H(z)$ implied by several properties of the system under concern. Using some of them helps in this question.

Prop 1: Total number of poles equals total number of zeros (including those at $z=0$ and $|z|=\infty$) ,i.e., There are equal number of poles and zeros for any $H(z)$.

Rule 2: For real systems (which has real impulse response $h[n]$) poles (or zeros) exist as either purely real or as complex conjugate pairs.

Now applying these two properties to your system which is real and causal (how?) we reach the following conclusion:

1- If ROC of $(1-\frac{1}{2} z^{-1}) H(z)$ is the entire $\mathcal{Z}$-plane (exc $z=0$), then $H(z)$ has a pole at $z=0.5$

2- If $x[n]=(1+j)^n$ produces zero output, then $z=1 \pm j$ are two zeros of $H(z)$

Are there any more poles of $H(z)$ other than those at $z=0$ ? No. Because if there were so, then part one would indicate it in the ROC of $(1-0.5 z^{-1})H(z)$ which is given as entire $\mathcal{Z}$ plane indicating that there are no more poles of finite, nonzeros values. 

Are there any more zeros of $H(z)$ ? The given information is not sufficient to conclude that. There could be more zeros. If they were real then they could be even or odd numbered, if they were complex then they must be in complex conjugate pairs. We only know one complex pair of zeros at $z=1 \pm j$. Note that for each added zero to $H(z)$ there would be one more added pole at $z=0$ (in addition to the existing pole at $z=0$)

Hence we conclude: $H(z)$ has two poles at $z=0$ and $z=0.5$ and at least two zeros at $z=1+j$ and $z=1-j$. If $H(z)$ had any more zeros at $z_k=\alpha_k$ then there would be equal number of poles at $z=0$. And if the zeros were complex they would be in complex-conjugate pairs (and the added poles would be doubled)