解：$(1)$由题意可知，该饮料的体积$V_{饮料}=V_{浓}+V_{水}，$即$250\ \mathrm {mL}=250\ \mathrm {cm}^3=V_{浓}+V_{水} ①，$该饮料的质量$m_{饮料}=m_{浓}+m_{水}=ρ_{浓}V_{浓}+ρ_{水}V_{水}，$即$262\ \mathrm {g}=1.6\ \mathrm {g/cm}^3×V_{浓}+1.0\ \mathrm {g/cm}^3×V_{水} ②，$由$①②$解得加入浓缩橙汁的体积$V_{浓}=20\ \mathrm {mL}，$加入水的体积$V_{水}=230\ \mathrm {mL}；$该饮料浓缩橙汁含量为$\frac {V_{浓}}{V_{饮料}}=\frac {20\ \mathrm {mL}}{250\ \mathrm {mL}}×100\%=8\%，$测得的浓缩橙汁含量小于$10\%，$不符合配料表所写
$(2)$配料表标明了浓缩橙汁含量$≥10\%，$设加入的浓缩橙汁的体积为$V_{加}，$则有$\frac {V_{浓}+V_{加}}{V_{饮料}+V_{加}}×100\%=10\%，$即$\frac {20\ \mathrm {mL}+V_{加}}{250\ \mathrm {mL}+V_{加}}×100\%=10\%，$解得：$V_{加}≈5.6\ \mathrm {mL}，$故这杯饮料中至少还要添加$5.6\ \mathrm {mL} $浓缩橙汁，才能达到要求
